poj 初期基本搜索

 

第四个专题了，初期基本搜索:

都是水题，两天完全可以刷完。。。

(1)、深度优先搜索

1、poj2488

题意：给出一个国际棋盘的大小，判断马能否不重复的走过所有格，并记录下其中按字典序排列的第一种路径。

分析：爆搜。。。

#include <iostream>
#include <cstring>
using namespace std;

const int N = 35;
int vis

;
int p, q, flag;
char b[500];
int dir[8][2] = {-1, -2, 1, -2, -2, -1, 2, -1, -2, 1, 2, 1, -1, 2, 1, 2};

void dfs(int x, int y, int cnt) {
if (cnt == p*q) {
cout << "A1";
for (int i = 2; i < 2*cnt; i++) cout << b[i];
cout << endl << endl;
flag = 1;
return;
}
for (int i = 0; i < 8 && !flag; i++) {
int tx = x + dir[i][0], ty = y + dir[i][1];
if (tx < 1 || tx > p || ty < 1 || ty > q) continue;
if (!vis[tx][ty]) {
vis[tx][ty] = 1;
b[2*cnt] = 'A' + ty - 1;
b[2*cnt+1] = '1' + tx - 1;
dfs(tx, ty, cnt+1);
vis[tx][ty] = 0;
}
}
}

int main() {
int t, ca = 0;
cin >> t;
while (t--) {
cin >> p >> q;
cout << "Scenario #" << ++ca << ':' << endl;
memset(vis, 0, sizeof(vis));
flag = 0;
vis[1][1] = 1;
dfs(1, 1, 1);
if (!flag) cout << "impossible" << endl << endl;
}
return 0;
}

2、poj3083

题意：给定一个迷宫，S是起点，E是终点，#是墙不可走，.可以走先输出左转优先时，从S到E的步数，再输出右转优先时，从S到E的步数，最后输出S到E的最短步数

分析：dfs找左转和右转步数，bfs找最短路。

 

#include <cstdio>
#include <cstring>

int w, h, c, a;
int Sx, Sy, Ex, Ey;
int cnt, flag, tag;
int dir[4][2] = {0, -1, -1, 0, 0, 1, 1, 0};
char maze[50][50];
int vis[50][50];
int que[2500][2];

void DFS(int x, int y, int cnt) {
if (maze[x][y] == 'E') {
printf("%d ", cnt);
tag = 0;
return ;
}
if (tag) for (int i = a+c, j = 0; j < 4; j++, i -= c) {
if (i < 0) i = 3;
if (i > 3) i = 0;
int tx = x + dir[i][0];
int ty = y + dir[i][1];
if (tx > h || tx < 1 || ty > w || ty < 1) continue;
if (tag && !vis[tx][ty] && maze[tx][ty] != '#') {
a = i;
DFS(tx, ty, cnt+1);
}
}
}

void BFS() {
int fir = 0, sec = 0;
que[sec][0] = Sx;
que[sec++][1] = Sy;
int step = 1;
while(fir < sec && !vis[Ex][Ey]) {
int tmp = sec;
step++;
while(fir < tmp && !vis[Ex][Ey]) {
int x = que[fir][0];
int y = que[fir++][1];
for(int i = 0; i < 4; i++) {
int tx = x + dir[i][0];
int ty = y + dir[i][1];
if (tx > h || tx < 1 || ty > w || ty < 1) continue;
if(!vis[tx][ty] && maze[tx][ty] != '#') {
que[sec][0] = tx;
que[sec++][1] = ty;
vis[tx][ty] = 1;
}
}
}
}
printf("%d\n", step);
}

int main() {
int t, i, j;
scanf("%d", &t);
while (t--) {
memset(maze, 0, sizeof(maze));
memset(vis, 0, sizeof(vis));
scanf("%d %d", &w, &h);
getchar();
for (i = 1; i <= h; i++) {
for (j = 1; j <= w; j++) {
scanf("%c", &maze[i][j]);
if (maze[i][j] == 'S') Sx = i, Sy = j;
else if (maze[i][j] == 'E') Ex = i, Ey = j;
}
getchar();
}
for (i = 0; i < 4; i++)
if (maze[Sx+dir[i][0]][Sy+dir[i][1]] == '.') break;
vis[Sx][Sy] = 1;
a = i; tag = 1; c = -1;
DFS(Sx, Sy, 1);
a = i; c = 1; tag = 1;
DFS(Sx, Sy, 1);
BFS();
}
return 0;
}

3、poj3009

题意：要求把一个冰壶从起点“2”用最少的步数移动到终点“3”，其中0为移动区域，1为石头区域，冰壶一旦想着某个方向运动就不会停止，也不会改变方向（想想冰壶在冰上滑动），除非冰壶撞到石头1或者到达终点3，冰壶撞到石头后，冰壶会停在石头前面，此时（静止状态）才允许改变冰壶的运动方向，而该块石头会破裂，石头所在的区域由1变为0. 也就是说，冰壶撞到石头后，并不会取代石头的位置。终点是一个摩擦力很大的区域，冰壶若到达终点3，就会停止在终点的位置不再移动。

分析：直接搜索，题目细节是沿着某一行出发会遇到石头才会停止，不停止则失败，停止时石头也会消失，步数不能超过10步。

#include <cstdio>
using namespace std;

int square[30][30];
int w, h, sx, sy;
int minnum;
int dir[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};

void DFS(int x, int y, int cnt) {
if (cnt >= 10) return ;
for (int i = 0; i < 4; i++) {
int flag = 1, tag = 0;
int nx = x + dir[i][0], ny = y + dir[i][1];
while (1) {
if (nx < 1 || nx > h || ny < 1 || ny > w) {
flag = 0;
break;
}
if (square[nx][ny] == 1) {
tag = 1;
break;
}
if (square[nx][ny] == 3 && minnum > ++cnt) {
minnum = cnt;
return ;
}
nx += dir[i][0];
ny += dir[i][1];
}
if (!flag) continue;
if (tag) {
square[nx][ny] = 0;
nx -= dir[i][0];
ny -= dir[i][1];
if (nx == x && ny == y) {
square[nx+dir[i][0]][ny+dir[i][1]] = 1;
continue;
}
DFS(nx, ny, cnt+1);
square[nx+dir[i][0]][ny+dir[i][1]] = 1;
}
}
}

int main() {
while (scanf("%d %d", &w, &h), w+h) {
minnum = 9999;
for (int i = 1; i <= h; i++) {
for (int j = 1; j <= w; j++){
scanf("%d", &square[i][j]);
if (square[i][j] == 2){
sx = i;
sy = j;
}
}
}
DFS(sx, sy, 0);
minnum <= 10 ? printf("%d\n", minnum) : puts("-1");
}
return 0;
}

4、poj1321

题意：中文题。。。

分析：在n*n的格子中选取k列来放棋子，每次确定列数之后在确定行数，直接搜索。

#include <cstdio>
#include <cstring>
using namespace std;

const int N = 12;
int n, k;
char b

;
bool visr
, visc
;

void init() {
memset(visr, 0, sizeof(visr));
memset(visc, 0, sizeof(visc));
}
int dfs(int col, int cnt) {
if (cnt == k) return 1;
int t = n - k + cnt + 1, res  = 0;
for (int j = col; j < t; j++) {
for (int i = 0; i < n; i++) {
if (b[i][j] == '#' && !visc[j] && !visr[i]) {
visr[i] = visc[j] = 1;
res += dfs(j+1, cnt+1);
visr[i] = visc[j] = 0;
}
}
}
return res;
}
int main() {
while (scanf("%d %d", &n, &k), n != -1) {
init();
for (int i = 0; i < n; i++) scanf("%s", b[i]);
printf("%d\n", dfs(0, 0));
}
return 0;
}

5、poj2251

题意：一个立体空间，有L个平面，R行，C列，.代表可以走，#代表不能走，S代表开始点，E代表结束点，问从S走到E点，最少可以经过多少分钟，若不能到达，则输出Trapped!

分析：bfs水题，三维的，没什么变化，多一种行走方式而已。

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int N = 35;
const int INF = 0x3f3f3f3f;
char dun

;
bool vis

;
int l, r, c;
int dx[] = {1, -1, 0, 0, 0, 0}, dy[] = {0, 0, 1, -1, 0, 0}, dz[] = {0, 0, 0, 0, 1, -1};

struct po {
int x, y, z, step;
po(int x, int y, int z, int s) : x(x), y(y), z(z), step(s) {}
po() {}
}s, e;

queue<po> q;
void init() {
while (!q.empty()) q.pop();
q.push(s);
for (int i = 0;  i < l; i++)
for (int j = 0; j < r; j++) memset(vis[i][j], 0, 4*(c+3));
}
int bfs() {
init();
while (!q.empty()) {
po t = q.front(); q.pop();
if (t.x == e.x && t.y == e.y && t.z == e.z) return t.step;
for (int i = 0; i < 6; i++) {
int nx = t.x + dx[i], ny = t.y + dy[i], nz = t.z + dz[i];
if (nx >= 0 && nx < l && ny >= 0 && ny < r && nz >= 0 &&
nz < c && !vis[nx][ny][nz] && dun[nx][ny][nz] != '#') {
q.push(po(nx, ny, nz, t.step+1)); vis[nx][ny][nz] = 1;
}
}
}
return INF;
}
int main() {
while (~scanf("%d %d %d", &l, &r, &c), l+r+c) {
int flags = 1, flage = 1;
for (int i = 0; i < l; i++) {
for (int j = 0; j < r; j++) scanf("%s", dun[i][j]);
for (int j = 0;  j < r && flags; j++)
for (int k = 0; k < c; k++) {
if (dun[i][j][k] == 'S') s = po(i, j, k, 0), flags = 0;
}
for (int j = 0;  j < r && flage; j++)
for (int k = 0; k < c; k++) {
if (dun[i][j][k] == 'E') e = po(i, j, k, 0), flage = 0;
}
}
int ans = bfs();
ans < INF ? printf("Escaped in %d minute(s).\n", ans) : puts("Trapped!");
}
return 0;
}

 (2)、广度优先搜索

1、poj3278

题意：就是给出a和b点的横坐标，求到a，b的最小行动次数，每次可以向左或向右移动一步，即横坐标加1或者减1，也可以横坐标变成原来的两倍。

分析：bfs的最基本应用，找最小步数/最小时间等。

#include <cstdio>
#include <queue>
#define MAXN 100000

using namespace std;
int N, K;
int vis[MAXN+1];

struct po { int p, time; };
queue <po> q;
void BFS() {

po temp = {N, 0};
q.push(temp);
vis
= 1;
while (!q.empty()) {
int a = q.front().p, b = q.front().time;
q.pop();
if (a == K) {
printf("%d\n", b);
return ;
}
if (a+1 <= MAXN && !vis[a+1]) {
temp.p = a+1;
temp.time = b+1;
q.push(temp);
vis[a+1] = 1;
}
if (a-1 >= 0 && !vis[a-1]) {
temp.p = a-1;
temp.time = b+1;
q.push(temp);
vis[a-1] = 1;
}
if (2*a <= MAXN && !vis[2*a]) {
temp.p = 2*a;
temp.time = b+1;
q.push(temp);
vis[2*a] = 1;
}
}
}

int main() {
scanf("%d %d", &N, &K);
BFS();
return 0;
}

2、poj1426

题意：给定一个数，找到由一个01串使得是它的倍数。

分析：直接bfs每次加0加1，同时每次取模，判断是不是为0就可以了。

#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

const int N = 1000000;
int flag, mod, cnt, i, j;
int prior
, prbit
;
char s[110];
int next[2] = {1, 0};

struct mul { int num, prev; };

struct Queue {
mul q[N*10];
int front, rear;
};
Queue Q;

void BFS() {
int a, b, c, d;
flag = cnt = 1;
mul sta = {1, 1};
Q.front = Q.rear = 0;
Q.q[Q.rear] = sta;
while (flag) {
a = Q.q[Q.front].num;
b = Q.q[Q.front++].prev;
for (j = 0; j < 2 && flag; j++) {
prior[++cnt] = b;
prbit[cnt] = next[j];
c = (a*10 + next[j]) % mod;
if (c == 0) {
i = 0;
d = cnt;
while (d != 1) {
s[i++] = prbit[d]+'0';
d = prior[d];
}
s[i++] = '1';
flag = 0;
break;
}
mul temp = {c, cnt};
Q.q[++Q.rear] = temp;
}
}
}

int main() {
while (scanf("%d", &mod), mod) {
memset(s, 0, sizeof(s));
BFS();
reverse(s, s+i);
puts(s);
}
return 0;
}

3、poj3126

题意：给定两个素数a b，求a变换到b需要几步，并且变换时只有一个数字不同。

分析：素数打表，直接bfs。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <utility>
#include <string>
using namespace std;
int prime[] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,1597,1601,1607,1609,1613,1619,1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,1973,1979,1987,1993,1997,1999,2003,2011,2017,2027,2029,2039,2053,2063,2069,2081,2083,2087,2089,2099,2111,2113,2129,2131,2137,2141,2143,2153,2161,2179,2203,2207,2213,2221,2237,2239,2243,2251,2267,2269,2273,2281,2287,2293,2297,2309,2311,2333,2339,2341,2347,2351,2357,2371,2377,2381,2383,2389,2393,2399,2411,2417,2423,2437,2441,2447,2459,2467,2473,2477,2503,2521,2531,2539,2543,2549,2551,2557,2579,2591,2593,2609,2617,2621,2633,2647,2657,2659,2663,2671,2677,2683,2687,2689,2693,2699,2707,2711,2713,2719,2729,2731,2741,2749,2753,2767,2777,2789,2791,2797,2801,2803,2819,2833,2837,2843,2851,2857,2861,2879,2887,2897,2903,2909,2917,2927,2939,2953,2957,2963,2969,2971,2999,3001,3011,3019,3023,3037,3041,3049,3061,3067,3079,3083,3089,3109,3119,3121,3137,3163,3167,3169,3181,3187,3191,3203,3209,3217,3221,3229,3251,3253,3257,3259,3271,3299,3301,3307,3313,3319,3323,3329,3331,3343,3347,3359,3361,3371,3373,3389,3391,3407,3413,3433,3449,3457,3461,3463,3467,3469,3491,3499,3511,3517,3527,3529,3533,3539,3541,3547,3557,3559,3571,3581,3583,3593,3607,3613,3617,3623,3631,3637,3643,3659,3671,3673,3677,3691,3697,3701,3709,3719,3727,3733,3739,3761,3767,3769,3779,3793,3797,3803,3821,3823,3833,3847,3851,3853,3863,3877,3881,3889,3907,3911,3917,3919,3923,3929,3931,3943,3947,3967,3989,4001,4003,4007,4013,4019,4021,4027,4049,4051,4057,4073,4079,4091,4093,4099,4111,4127,4129,4133,4139,4153,4157,4159,4177,4201,4211,4217,4219,4229,4231,4241,4243,4253,4259,4261,4271,4273,4283,4289,4297,4327,4337,4339,4349,4357,4363,4373,4391,4397,4409,4421,4423,4441,4447,4451,4457,4463,4481,4483,4493,4507,4513,4517,4519,4523,4547,4549,4561,4567,4583,4591,4597,4603,4621,4637,4639,4643,4649,4651,4657,4663,4673,4679,4691,4703,4721,4723,4729,4733,4751,4759,4783,4787,4789,4793,4799,4801,4813,4817,4831,4861,4871,4877,4889,4903,4909,4919,4931,4933,4937,4943,4951,4957,4967,4969,4973,4987,4993,4999,5003,5009,5011,5021,5023,5039,5051,5059,5077,5081,5087,5099,5101,5107,5113,5119,5147,5153,5167,5171,5179,5189,5197,5209,5227,5231,5233,5237,5261,5273,5279,5281,5297,5303,5309,5323,5333,5347,5351,5381,5387,5393,5399,5407,5413,5417,5419,5431,5437,5441,5443,5449,5471,5477,5479,5483,5501,5503,5507,5519,5521,5527,5531,5557,5563,5569,5573,5581,5591,5623,5639,5641,5647,5651,5653,5657,5659,5669,5683,5689,5693,5701,5711,5717,5737,5741,5743,5749,5779,5783,5791,5801,5807,5813,5821,5827,5839,5843,5849,5851,5857,5861,5867,5869,5879,5881,5897,5903,5923,5927,5939,5953,5981,5987,6007,6011,6029,6037,6043,6047,6053,6067,6073,6079,6089,6091,6101,6113,6121,6131,6133,6143,6151,6163,6173,6197,6199,6203,6211,6217,6221,6229,6247,6257,6263,6269,6271,6277,6287,6299,6301,6311,6317,6323,6329,6337,6343,6353,6359,6361,6367,6373,6379,6389,6397,6421,6427,6449,6451,6469,6473,6481,6491,6521,6529,6547,6551,6553,6563,6569,6571,6577,6581,6599,6607,6619,6637,6653,6659,6661,6673,6679,6689,6691,6701,6703,6709,6719,6733,6737,6761,6763,6779,6781,6791,6793,6803,6823,6827,6829,6833,6841,6857,6863,6869,6871,6883,6899,6907,6911,6917,6947,6949,6959,6961,6967,6971,6977,6983,6991,6997,7001,7013,7019,7027,7039,7043,7057,7069,7079,7103,7109,7121,7127,7129,7151,7159,7177,7187,7193,7207,7211,7213,7219,7229,7237,7243,7247,7253,7283,7297,7307,7309,7321,7331,7333,7349,7351,7369,7393,7411,7417,7433,7451,7457,7459,7477,7481,7487,7489,7499,7507,7517,7523,7529,7537,7541,7547,7549,7559,7561,7573,7577,7583,7589,7591,7603,7607,7621,7639,7643,7649,7669,7673,7681,7687,7691,7699,7703,7717,7723,7727,7741,7753,7757,7759,7789,7793,7817,7823,7829,7841,7853,7867,7873,7877,7879,7883,7901,7907,7919,7927,7933,7937,7949,7951,7963,7993,8009,8011,8017,8039,8053,8059,8069,8081,8087,8089,8093,8101,8111,8117,8123,8147,8161,8167,8171,8179,8191,8209,8219,8221,8231,8233,8237,8243,8263,8269,8273,8287,8291,8293,8297,8311,8317,8329,8353,8363,8369,8377,8387,8389,8419,8423,8429,8431,8443,8447,8461,8467,8501,8513,8521,8527,8537,8539,8543,8563,8573,8581,8597,8599,8609,8623,8627,8629,8641,8647,8663,8669,8677,8681,8689,8693,8699,8707,8713,8719,8731,8737,8741,8747,8753,8761,8779,8783,8803,8807,8819,8821,8831,8837,8839,8849,8861,8863,8867,8887,8893,8923,8929,8933,8941,8951,8963,8969,8971,8999,9001,9007,9011,9013,9029,9041,9043,9049,9059,9067,9091,9103,9109,9127,9133,9137,9151,9157,9161,9173,9181,9187,9199,9203,9209,9221,9227,9239,9241,9257,9277,9281,9283,9293,9311,9319,9323,9337,9341,9343,9349,9371,9377,9391,9397,9403,9413,9419,9421,9431,9433,9437,9439,9461,9463,9467,9473,9479,9491,9497,9511,9521,9533,9539,9547,9551,9587,9601,9613,9619,9623,9629,9631,9643,9649,9661,9677,9679,9689,9697,9719,9721,9733,9739,9743,9749,9767,9769,9781,9787,9791,9803,9811,9817,9829,9833,9839,9851,9857,9859,9871,9883,9887,9901,9907,9923,9929,9931,9941,9949,9967,9973,0};
typedef pair<string, int> p;
queue<p> q;
int d[4][12] = {{9, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {10, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9},
{10, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {4, 1, 3, 7, 9}};
int vis[10000], isprime[10000];

int solve(string f, string s) {
while (!q.empty()) q.pop();
memset(vis, 0, sizeof(vis));
q.push(p(f, 0));
while (!q.empty()) {
p t = q.front(); q.pop();
string x = t.first;
if (x == s) return t.second+1;
for (int i = 0;  i< 4; i++) {
for (int j = 1; j <= d[i][0]; j++) {
string tmp = x;
tmp[i] = d[i][j] + '0';
int k = atoi(tmp.c_str());
if (!vis[k] && isprime[k]) vis[k] = 1, q.push(p(tmp, t.second+1));
}
}
}
return 0;
}
int main() {
for (int i = 0; prime[i]; i++) isprime[prime[i]] = 1;
int t;
scanf("%d", &t);
while (t--) {
char f[10], s[10];
scanf("%s %s", &f, &s);
int ans = solve(f, s);
vis[atoi(f)] = 1;
ans > 0 ? printf("%d\n", ans-1) : puts("Impossible");
}
return 0;
}

4、poj3087

题意：已知两堆牌s1和s2的初始状态， 其牌数均为c，按给定规则能将他们相互交叉组合成一堆牌s12，再将s12的最底下的c块牌归为s1，最顶的c块牌归为s2，依此循环下去。

现在输入s1和s2的初始状态以及 预想的最终状态s12。问s1 s2经过多少次洗牌之后，最终能达到状态s12，若永远不可能相同，则输出"-1"。

分析：直接模拟，set判重。

 

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <set>
using namespace std;

const int N = 210;
int len;
char f
, s
, last
;
string res, tt;
set<string> a;
void init() {
a.clear();
res = last; tt = "";
for (int i = 0; i < len; i++) tt += s[i], tt += f[i];
a.insert(tt);
}
int bfs() {
init();
int ans = 1;
while (1) {
if (tt == res) return ans;
string tmp = "";
for (int i = len; i < 2*len; i++) tmp += tt[i], tmp += tt[i-len];
if (a.count(tmp)) return -1;
else tt = tmp, a.insert(tt), ans++;
}
}
int main() {
int t, ca = 0;
scanf("%d", &t);
while (t--) {
scanf("%d", &len);
scanf("%s %s %s", f, s, last);
printf("%d %d\n", ++ca, bfs());
}
return 0;
}

5、Poj3414

题意：给出了两个瓶子的容量A,B, 以及一个目标水量C，

对A、B可以有如下操作:装满一个瓶子、倒空一个瓶子、一个瓶子倒水到另一个瓶子。

问经过哪几个操作后能使得任意一个瓶子的残余水量为C，且输出操作过程。若不可能得到则输出impossible。

分析：设几个状态区别操作，直接bfs，并且记录路径。

 

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;

const int N = 110;
bool vis

;
int a, b, c;
struct node {
int a, b, cur, step;
node(int a = 0, int b = 0, int c = 0, int s = 0) : a(a), b(b), cur(c), step(s) {}
};
queue<node> q;
int prior[N*N], ans[N*N];
void solve(int cur) {
if (prior[cur] == -1) return ;
solve(prior[cur]);
if (ans[cur] == 1) puts("DROP(1)");
else if (ans[cur] == 2) puts("FILL(1)");
else if (ans[cur] == 3) puts("POUR(1,2)");
else if (ans[cur] == 4) puts("DROP(2)");
else if (ans[cur] == 5) puts("FILL(2)");
else if (ans[cur] == 6) puts("POUR(2,1)");
}
bool bfs() {
int cnt = 0;
prior[0] = -1;
q.push(node());
while (!q.empty()) {
node t = q.front(); q.pop();
if (t.a == c || t.b == c) {
printf("%d\n", t.step);
solve(t.cur);
return 1;
}
int sum = t.a + t.b;
if (t.a) {
if (!vis[0][t.b]) {
q.push(node(0, t.b, ++cnt, t.step+1));
prior[cnt] = t.cur, vis[0][t.b] = 1;
ans[cnt] = 1;
}
int db = b - t.b;
if (t.a >= db && !vis[t.a-db]) {
q.push(node(t.a-db, b, ++cnt, t.step+1));
prior[cnt] = t.cur, vis[t.a-db][b] = 1;
ans[cnt] = 3;
}
if (t.a < db && !vis[0][sum]) {
q.push(node(0, sum, ++cnt, t.step+1));
prior[cnt] = t.cur; vis[0][sum] = 1;
ans[cnt] = 3;
}
}
else {
if (t.a != a && !vis[a][t.b]) {
q.push(node(a, t.b, ++cnt, t.step+1));
prior[cnt] = t.cur, vis[a][t.b] = 1;
ans[cnt] = 2;
}
}
if (t.b) {
if (!vis[t.a][0]) {
q.push(node(t.a, 0, ++cnt, t.step+1));
prior[cnt] = t.cur, vis[t.a][0] = 1;
ans[cnt] = 4;
}
int da = a - t.a;
if (t.b >= da && !vis[a][t.b-da]) {
q.push(node(a, t.b-da, ++cnt, t.step+1));
prior[cnt] = t.cur, vis[a][t.b-da] = 1;
ans[cnt] = 6;
}
if (t.b < da && !vis[sum][0]) {
q.push(node(sum, 0, ++cnt, t.step+1));
prior[cnt] = t.cur; vis[sum][0] = 1;
ans[cnt] = 6;
}
}
else {
if (b != t.b && !vis[t.a][b]) {
q.push(node(t.a, b, ++cnt, t.step+1));
prior[cnt] = t.cur, vis[t.a][b] = 1;
ans[cnt] = 5;
}
}
}
return 0;
}
int main() {
scanf("%d %d %d", &a, &b, &c);
if (!bfs()) puts("impossible");
return 0;
}

[b](3)、简单搜索技巧和剪枝

 

这里的几个题都可以直接搜，不需要剪枝，而且我感觉剪枝没多大用，因为本来效率就很高，不过可以学习最基本的一些剪枝技巧。

 

1、poj2531

题意：是把点分成A、B两组，节点之间的间距已给出，要求解分组的方法，使得∑Cij (i∈A,j∈B)最大

分析：直接暴力枚举A组，优化了下计算过程。不过这题归类为剪枝，当然也有剪枝的方式来搜索。方式同样是枚举A组的点，每次在A组中加入一个点，计算新的权值，如果答案更大，则把该点加入A集合中，继续进行枚举，否则就没必要继续进行枚举了。枚举新加入的其他点即可。

 

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <numeric>
using namespace std;

const int N = 25;
int c

, sum
, tmp
, val[1<<21], a
;
int ans = 0, n, up;

void dfs(int i, int* a, int k, int t) {
if (i == n) {
if (val[t] == -1) {
int tot = 0;
for (int i = 0; i < k; i++) {
tot += sum[a[i]];
for (int j = 0; j < k; j++) tot -= c[a[i]][a[j]];
}
val[t] = val[up^t] = tot;
ans = max(ans, tot);
}
return ;
}
dfs(i+1, a, k, t);
a[k++] = i;
dfs(i+1, a, k, t | tmp[i]);
}
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) tmp[i] = 1 << i;
up = (1 << n) - 1;
memset(val, -1, 4*(up+10));
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) scanf("%d", &c[i][j]);
for (int i = 0; i < n; i++) sum[i] = accumulate(c[i], c[i]+n, 0);
dfs(0, a, 0, 0);
printf("%d\n", ans);
return 0;
}

2、poj1416

题意：将一个数字m（不超过六位）剪成几段，要求得到的新的几个数之和不超过另外一个给定的数n，求可以得到的最接近n的数。

分析：简单的搜索题，随便暴力都能过，剪枝的方式就是如果当前枚举出的切割方式使得和已经比目标大了，则没必要继续搜索。这是可行性剪枝。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;

const int N = 10;
int dig
, tmp
, ans
;
int tar, a, len, res, cnt, l;

void dfs(int i, int sum, int* a, int k) {
if (sum > tar) return ;
if (i > len) {
if (sum == res) cnt++;
else if (sum > res) {
res = sum; l = k;
memcpy(ans, a, 4*k), cnt = 1;
}
}
for (int j = i; j <= len; j++) {
int val = 0;
for (int x = i; x <= j; x++) val = val * 10 + dig[x];
a[k] = val;
dfs(j+1, sum + val, a, k+1);
}
}
int main() {
while (scanf("%d %d", &tar, &a), tar+a) {
if (tar == a) { printf("%d %d\n", tar, a); continue; }
int t = a, sum = 0;
len = log10(a) + 1;
for (int i = len; i >= 1; i--, t /= 10) dig[i] = t % 10, sum += dig[i];
if (sum > tar) { puts("error"); continue; }
cnt = res = 0;
dfs(1, 0, tmp, 0);
if (cnt > 1) puts("rejected");
else {
printf("%d", res);
for (int j = 0; j < l; j++) printf(" %d", ans[j]);
puts("");
}
}
return 0;
}

3、poj2676

题意：把一个9行9列的网格，再细分为9个3*3的子网格，要求每行、每列、每个子网格内都只能使用一次1~9中的一个数字，即每行、每列、每个子网格内都不允许出现相同的数字。求满足题意的数独。

分析：直接搜索，标记行、列、块，值得一提的是倒着搜比正着搜效率高出许多，这也算是一个技巧。

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int N = 25;
int a

;
char s[10];
int k, flag = 1;
struct po{int x, y;}p[90];
int visro

, visco

, visbl[10]
;
po pos[] = {{0, 0}, {1, 1}, {1, 4}, {1, 7}, {4, 1}, {4, 4}, {4, 7}, {7, 1}, {7, 4}, {7, 7}};
int num[10][10];

void DFS(int cnt) {
if (cnt < 0) {
for (int i = 1; i <= 9; i++){
for (int j = 1; j <= 9; j++) printf("%d", a[i][j]);
putchar('\n');
}
flag = 0;
return ;
}
if (flag){
int x = p[cnt].x, y = p[cnt].y;
for (int i = 1; i <= 9 && flag; i++){
if (!visco[y][i] && !visro[x][i] && !visbl[num[x][y]][i]) {
visco[y][i] = visro[x][i] = visbl[num[x][y]][i] = 1;
a[x][y] = i;
DFS(cnt-1);
if (!flag) break;
visco[y][i] = visro[x][i] = visbl[num[x][y]][i] = 0;
}
}
}
}
int main() {
for (int i = 1; i <= 9; i++) {
int x = pos[i].x, y = pos[i].y;
for (int j = 0; j < 3; j++)
for (int l = 0; l < 3; l++) num[x+j][y+l] = i;
}
int t;
scanf("%d", &t);
while (t--) {
flag = 1; k = 0;
for (int i = 1; i <= 9; i++) {
scanf("%s", s+1);
for (int j = 1; j <= 9; j++) {
a[i][j] = s[j]-'0';
if (!a[i][j]) p[k++] = (po){i, j};
}
}
memset(visro, 0, sizeof(visro));
memset(visco, 0, sizeof(visco));
memset(visbl, 0, sizeof(visbl));
for (int i = 1; i <= 9; i++) {
for (int j = 1; j <= 9; j++) visro[i][a[i][j]] = visco[i][a[j][i]] = 1;
int x = pos[i].x, y = pos[i].y;
for (int j = 0; j < 3; j++)
for (int l = 0; l < 3; l++) visbl[i][a[x+j][y+l]] = 1;
}
DFS(k-1);
}
return 0;
}

4、poj1129

题意：给一个图，相邻的结点关系已给出，求给图着色要多少种颜色。

分析：直接暴力枚举，颜色数从小到大枚举即可，不过这题简直神坑，刚开始我读错题了，以为是枚举全排列，存在相邻关系的点在排列中相邻会多一种颜色，然后在那各种剪枝，但是14以上的数据基本上不可能在时限内解出来了。。。交上去居然没超时，折腾了一阵发现读错题了，重写了一遍，这题无语主要是数据过弱，网上很多错误代码都过了，我跑数据测试过。难怪我枚举全排列都没超时。。。。

#include <cstdio>
#include <cstring>
using namespace std;

const int N = 30;
bool g

;
char s[N<<1];
int val
, n;

bool cal(int x, int v) {
for (int i = 0; i < n; i++)
if (g[x][i] && val[i] == v) return 0;
return 1;
}
bool dfs(int x, int num) {
if (x == n) return 1;
for (int i = 1; i <= num; i++) {
if (cal(x, i)) {
val[x] = i;
if (dfs(x+1, num)) return 1;
}
}
return 0;
}
int main() {
while (scanf("%d", &n), n) {
for (int i = 0; i < n; i++) {
memset(g[i], 0, 4*n);
val[i] = 0;
scanf("%s", s);
for (int j = 2; s[j]; j++) g[i][s[j]-'A'] = 1;
}
int ans = 0;
for (int i = 1; !ans; i++)
if (dfs(0, i)) ans = i;
ans == 1 ? puts("1 channel needed.") : printf("%d channels needed.\n", ans);
}
return 0;
}

总结：第四个专题是最基本的搜索题目训练，题目都很简单，可以帮助巩固基础。至于后面的剪枝部分的四个题目，也算是体现了一些基本的剪枝策略。

大概看来可以得出以下几点：

1、 搜索顺序很重要，比如说数独那题，这题倒着来效率比正着来效率更高，正是体现了搜索的顺序，有时我们还需要对搜索对象做些处理，原则是使得搜索更高效，更易剪枝，比如说处理搜索对象使得更易得到最优解，进而进行最优性剪枝，把不灵活的搜索对象放在前面可以进行可行性剪枝等。

2、 可行性剪枝，比如说切纸中如果和已经比目标大了，直接返回。

3、 最优性剪枝，第一个题的搜索方式倒是体现了一点，如果当前加入的点不可以得到更大的最优解直接退出。

我们需要的是根据以上三点得出更高效的搜索方式。

以上只是个人对剪枝的一点点看法，也不是什么经验之谈，望大牛不吝赐教。