http://acm.hdu.edu.cn/showproblem.php?pid=5040

一个人拿着纸盒子往目的地走  正常情况下一秒走一格  可以原地不动躲在盒子里  也可以套着盒子三秒走一格

地图上有些灯  灯能照到自己和面前一个格  每一秒灯顺时针转90度  如果要从灯照的地方离开或者进入灯照的地方就必须套上盒子

最短时间到达

题意不清的bfs

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <bitset>
#include <queue>
#include <iostream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define clr0(x) memset(x,0,sizeof(x))
typedef long long LL;
const int maxn = 505;
int sx,sy,n;
int dx[] = {0,1,0,-1},dy[] = {1,0,-1,0};
char s[maxn][maxn];
int dirr[128],notic[maxn][maxn];
bool vis[maxn][maxn][4];
bool in(int x,int y)
{return 0 <= x && x < n && 0 <= y && y < n;
}
struct node{int t,x,y;bool operator < (const node &a)const{return t > a.t;}
};
int bfs()
{priority_queue <node> q;q.push((node){0,sx,sy});while(!q.empty()){node now = q.top(),to;q.pop();if(s[now.x][now.y] == 'T'){return now.t;}if(vis[now.x][now.y][now.t%4])   continue;vis[now.x][now.y][now.t%4] = true;to = now,to.t++;q.push(to);for(int i = 0;i < 4;++i){int mx = now.x + dx[i],my = now.y + dy[i];if(in(mx,my) && s[mx][my] != '#'){//所在格子和目的格子同一秒没有摄像头的时候才能走to.t = now.t + 1;if( (notic[mx][my] | notic[now.x][now.y]) & (1<<(now.t%4)) )to.t = now.t + 3;to.x = mx,to.y = my;q.push(to);}}}return -1;
}
int main (){int  _,cas = 1;RD(_);dirr['E'] = 0,dirr['S'] = 1,dirr['W'] = 2,dirr['N'] = 3;dirr['T'] = dirr['M'] = dirr['.'] = dirr['#'] = -1;while(_--){printf("Case #%d: ",cas++);RD(n);clr0(notic);clr0(vis);for(int i = 0;i < n;++i){scanf("%s",s[i]);for(int j = 0;j < n;++j){if(s[i][j] == 'M')sx = i,sy = j;else{int now = dirr[ s[i][j] ];if(now == -1)continue;notic[i][j] = (1<<4) - 1;for(int k = now;k < 4+now;++k){int mx = i + dx[k%4],my = j + dy[k%4];if(in(mx,my)){notic[mx][my] |= (1<<(k-now));}}}}}cout<<bfs()<<endl;}return 0;
}