hdu 1495 非常可乐 (bfs)

小记：这题给我留下了一个TLE的印记，之前因为用的标记数组是int型的，导致了一直TLE，唉，memset一个int型数组浪费的时间真多，不然我之前的那个没用结构体的直接bfs的应该可以A过去，唉，又，b了。一般在写代码前都会先分析自己的算法的时间复杂度的大小，而觉得可以的，动手写了之后还是超时，要么就是你的想法错了，要么就是你的优化剪枝不够，而对于常数级优化 则是没必要的。谨记。。。

题解：题目要求 n + m == s 那么每一次的倒可乐就可以从这一杯倒到另外两杯去，而有三个杯子因此有6种倒的方法。即相当于一颗六叉树。bfs 搜索每一种状态，当有两个杯子是s的一半的时候，那个状态就是答案，输出即可。已搜过的进行标记。

剪枝：若s 为奇数，则无解，分不了相等的两半，因为你不能倒出一个0.5的状态值。

优化：标记数组设为bool型。我的设int型，156MS，设bool型，31MS。

代码奉上:

#include <stdio.h>
#include <string.h>
#include <queue>

using namespace std;
#define MAX 101

struct NODE {
int x,y,z,t;
};

bool visited[MAX][MAX][MAX];

void bfs(int s,int n,int m) {
queue<NODE>q;
NODE now, next;
now.x = s;
now.y = 0;
now.z = 0;
now.t = 0;
q.push(now);
memset(visited,0,sizeof(visited));
while(!q.empty()) {
now = q.front();
q.pop();
if((now.x == s>>1 && now.y == s>>1) || (now.x == s>>1 && now.z == s>>1) || (now.z == s>>1 && now.y == s>>1)){
printf("%d\n",now.t);
return ;
}

if(now.x) {
if(now.x > (n - now.y)){
next.x = now.x - n + now.y;
next.y = n;
next.z = now.z;
}
else {
next.x = 0;
next.y = now.y + now.x;
next.z = now.z;
}

if(!visited[next.x][next.y][next.z]) {
visited[next.x][next.y][next.z] = 1;
next.t = now.t + 1;
q.push(next);
}

if(now.x > (m - now.z)){
next.x = now.x - m + now.z;
next.z = m;
next.y = now.y;
}
else {
next.x = 0;
next.z = now.z+ now.x;
next.y = now.y;
}
if(!visited[next.x][next.y][next.z]) {
visited[next.x][next.y][next.z] = 1;
next.t = now.t + 1;
q.push(next);
}
}

if(now.y) {
if(now.y > (s - now.x)){
next.y = now.y - s + now.x;
next.x = s;
next.z = now.z;
}
else {
next.y = 0;
next.x = now.y+ now.x;
next.z = now.z;
}
if(!visited[next.x][next.y][next.z]) {
visited[next.x][next.y][next.z] = 1;
next.t = now.t + 1;
q.push(next);
}

if(now.y > (m - now.z)){
next.y = now.y - m + now.z;
next.z = m;
next.x = now.x;
}
else {
next.y = 0;
next.z = now.z+ now.y;
next.x = now.x;
}
if(!visited[next.x][next.y][next.z]) {
visited[next.x][next.y][next.z] = 1;
next.t = now.t + 1;
q.push(next);
}
}

if(now.z) {
if(now.z > (s - now.x)){
next.z = now.z - s + now.x;
next.x = s;
next.y = now.y;
}
else {
next.z = 0;
next.x = now.z+ now.x;
next.y = now.y;
}

if(!visited[next.x][next.y][next.z]) {
visited[next.x][next.y][next.z] = 1;
next.t = now.t + 1;
q.push(next);
}
if(now.z > (n - now.y)){
next.z = now.z - n + now.y;
next.y = n;
next.x = now.x;
}
else {
next.z = 0;
next.y = now.z+ now.y;
next.x = now.x;
}
if(!visited[next.x][next.y][next.z]) {
visited[next.x][next.y][next.z] = 1;
next.t = now.t + 1;
q.push(next);
}
}
}
puts("NO");
return ;
}

int main() {
int s, n, m;
while(scanf("%d%d%d",&s,&n,&m),s||n||m) {
if(s%2){
puts("NO");
}
else {
bfs(s,n,m);
}
}
return 0;
}


简化一点点：

#include <stdio.h>
#include <string.h>
#include <queue>

using namespace std;
#define MAX 101

typedef struct n {
int x,y,z,t;
}NODE;

bool visited[MAX][MAX][MAX];
int s, n, m;

int judge(NODE p){
if((p.x == p.y && p.x + p.y == s) || (p.x == p.z && p.x + p.z == s) || (p.z == p.y && p.z + p.y == s)){
return 1;
}
return 0;
}

void bfs() {
queue<NODE>q;
NODE now,next;
now.x = s;
now.y = 0;
now.z = 0;
now.t = 0;
q.push(now);
memset(visited,0,sizeof(visited));
while(!q.empty()) {
now = q.front();
q.pop();
if(judge(now)){
printf("%d\n",now.t);
return ;
}

if(now.x) {
next.x = (now.x > (n - now.y))? now.x - n + now.y: 0;
next.y = (now.x > (n - now.y))? n: now.y + now.x;
next.z = now.z;

if(!visited[next.x][next.y][next.z]) {
visited[next.x][next.y][next.z] = 1;
next.t = now.t + 1;
q.push(next);
}

next.x = (now.x > (m - now.z))? now.x - m + now.z: 0;
next.z = (now.x > (m - now.z))? m: now.z + now.x;
next.y = now.y;
if(!visited[next.x][next.y][next.z]) {
visited[next.x][next.y][next.z] = 1;
next.t = now.t + 1;
q.push(next);
}
}

if(now.y) {
next.y = (now.y > (s - now.x))? now.y - s + now.x: 0;
next.x = (now.y > (s - now.x))? s: now.x + now.y;
next.z = now.z;

if(!visited[next.x][next.y][next.z]) {
visited[next.x][next.y][next.z] = 1;
next.t = now.t + 1;
q.push(next);
}
next.y = (now.y > (m - now.z))? now.y - m + now.z: 0;
next.z = (now.y > (m - now.z))? m: now.z + now.y;
next.x = now.x;
if(!visited[next.x][next.y][next.z]) {
visited[next.x][next.y][next.z] = 1;
next.t = now.t + 1;
q.push(next);
}
}

if(now.z) {
next.z = (now.z > (s - now.x))? now.z - s + now.x: 0;
next.x = (now.z > (s - now.x))? s: now.z + now.x;
next.y = now.y;

if(!visited[next.x][next.y][next.z]) {
visited[next.x][next.y][next.z] = 1;
next.t = now.t + 1;
q.push(next);
}

next.z = (now.z > (n - now.y))? now.z - n + now.y: 0;
next.y = (now.z > (n - now.y))? n: now.z + now.y;
next.x = now.x;
if(!visited[next.x][next.y][next.z]) {
visited[next.x][next.y][next.z] = 1;
next.t = now.t + 1;
q.push(next);
}
}
}
puts("NO");
return ;
}
int main() {
while(scanf("%d%d%d",&s,&n,&m),s||n||m) {
if(s%2){
puts("NO");
}
else {
bfs();
}
}
return 0;
}