您的位置:首页 > 其它

HDU 2128 Tempter of the Bone II BFS

2014-08-27 15:57 337 查看
状压整张图包括每个点的炸弹有没有被拿,墙壁有没有被炸。用优先队列存一下状态。

还有就是注意整数数溢出的问题。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <deque>
#include <bitset>
#include <list>
#include <cstdlib>
#include <climits>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <stack>
#include <sstream>
#include <numeric>
#include <fstream>
#include <functional>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int,int> pii;
const int INF = INT_MAX / 3;
const double eps = 1e-8;
const LL LINF = 1e17;
const double DINF = 1e60;
const int dx[] = {0,0,1,-1};
const int dy[] = {1,-1,0,0};
const int maxn = 8;
char mp[maxn][maxn];
int n,m,sx,sy,ex,ey;

struct Node {
int x,y,dis,bomb;
ULL maze,vis;
Node(int x,int y,ULL maze,int dis,int bomb,ULL vis):
x(x),y(y),maze(maze),dis(dis),bomb(bomb),vis(vis) {}
bool operator < (const Node &p) const {
if(dis != p.dis) return dis > p.dis;
int e_dis1 = abs(x - ex) + abs(y - ey);
int e_dis2 = abs(p.x - ex) + abs(p.y - ey);
if(e_dis1 != e_dis2) return e_dis1 > e_dis2;
if(bomb != p.bomb) return bomb < p.bomb;
return maze > p.maze;
}
};

struct st_Node {
int x,y,bomb,dis;
ULL maze,vis;
st_Node(Node &p) {
x = p.x; y = p.y; bomb = p.bomb;
maze = p.maze; vis = p.vis; dis = p.dis;
}

bool operator < (const st_Node &p) const {
if(x != p.x) return x < p.x;
if(y != p.y) return y < p.y;
if(bomb != p.bomb) return bomb < p.bomb;
if(maze != p.maze) return maze < p.maze;
return vis < p.vis;
}
};

set<st_Node> st;

inline int U(int x,int y) {
return x * m + y;
}

inline bool in_bd(int x,int y) {
return x >= 0 && x < n && y >= 0 && y < m;
}

inline ULL maze_comp() {
ULL ret = 0;
for(int i = 0;i < n;i++) {
for(int j = 0;j < m;j++) {
if(mp[i][j] == 'X') ret |= (1LL << U(i,j));
}
}
return ret;
}

inline bool insert_st(Node &tmp) {
st_Node now(tmp);
set<st_Node>::iterator ret = st.find(now);
if(ret == st.end()) {
st.insert(now); return true;
}
return false;
}

void printmap(ULL maze,int x,int y) {
for(int i = 0;i < n;i++) {
for(int j = 0;j < m;j++) {
if(i == x && j == y) printf("@");
else if(maze & (1LL << U(i,j))) printf("X");
else printf(".");
}
puts("");
}
}

void solve() {
st.clear();
priority_queue<Node> q;
q.push(Node(sx,sy,maze_comp(),0,0,0));
int ans = INF;
while(!q.empty()) {
Node now = q.top(); q.pop();
int x = now.x, y = now.y, dis = now.dis, bomb = now.bomb;
ULL maze = now.maze, vis = now.vis;
if(x == ex && y == ey) {
ans = dis;
break;
}
for(int i = 0;i < 4;i++) {
int nx = x + dx[i], ny = y + dy[i];
if(!in_bd(nx,ny)) continue;
if(maze & (1LL << U(nx,ny))) {
//如果是墙
if(bomb == 0) continue;
int nbomb = bomb - 1;
ULL nmaze = maze ^ (1LL << U(nx,ny));
Node nnode =
Node(nx,ny,nmaze,dis + 2,nbomb,vis);
if(insert_st(nnode)) q.push(nnode);
}
else {
//空地
if (mp[nx][ny] >= '1' && mp[nx][ny] <= '9' && !(vis & (1LL << U(nx,ny)))) {
//炸弹储备
ULL nvis = vis | (1LL << U(nx,ny));
int nbomb = bomb + mp[nx][ny] - '0';
Node nnode =
Node(nx,ny,maze,dis + 1,nbomb,nvis);
if(insert_st(nnode)) {
q.push(nnode);
}
}
else {
Node nnode = Node(nx,ny,maze,dis + 1,bomb,vis);
if(insert_st(nnode)) q.push(nnode);
}
}
}
}
printf("%d\n",ans == INF ? -1 : ans);
}

int main() {
while(scanf("%d%d",&n,&m), n) {
for(int i = 0;i < n;i++) {
for(int j = 0;j < m;j++) {
scanf(" %c",&mp[i][j]);
if(mp[i][j] == 'S') {
sx = i; sy = j; mp[i][j] = '.';
}
if(mp[i][j] == 'D') {
ex = i; ey = j; mp[i][j] = '.';
}
}
}
solve();
}
return 0;
}


  
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: