zoj1649Rescue

Rescue

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Time Limit: 1 Second      Memory Limit: 32768 KB

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Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)

Input

First line contains two integers stand for N and M.

Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.

Process to the end of the file.

Output

For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."

Sample Input

7 8
#.#####.
#.a#..r.
#..#x...
..#..#.#
#...##..
.#......
........

Sample Output

13

 

这是一道广度优先搜索的变形题，一般来说，广搜得到的距离都是步数最小，而这题，步数最小不代表距离最小；因此，设定数组mintime[n][n],当从原点经过某条路径到达（i,j）点所用的时间小于mintime[i-1][j-1](下标从0开始)，则更新mintime[i-1][j-1]的值，并将这一点入队……

 

#include<stdio.h>#include<string.h>#include<queue>using namespace std;const int Inf=1000000;struct point{ int x,y; int cost; int steps;};char map1[200][200];int mintime[200][200];int sx,sy,dx,dy;int m,n;int move[4][2]={{0,1},{0,-1},{1,0},{-1,0}};int BFS(point s){ int i; int fx,fy,fcost,fstep; queue<point> qu; qu.push(s); while(!qu.empty()) { point temp; temp=qu.front(); qu.pop(); for(i=0;i<4;i++) { fx=temp.x+move[i][0]; fy=temp.y+move[i][1]; if(fx>=0&&fx<m&&fy>=0&&fy<n&&map1[fx][fy]!='#') { fcost=temp.cost+1; fstep=temp.steps+1; if(map1[fx][fy]=='x') fcost++; if(fcost<mintime[fx][fy]) { mintime[fx][fy]=fcost; point t; t.x=fx;t.y=fy; t.cost=fcost;t.steps=fstep; qu.push(t); } } } } return mintime[dx][dy];};int main(){ int i,j; point start; while(scanf("%d%d",&m,&n)!=EOF) { memset(map1,0,sizeof(map1)); for(i=0;i<m;i++) scanf("%s",map1[i]); for(i=0;i<m;i++) for(j=0;j<n;j++) { mintime[i][j]=Inf; if(map1[i][j]=='r') { sx=i;sy=j; } if(map1[i][j]=='a') { dx=i;dy=j; } } start.x=sx;start.y=sy; start.steps=0;start.cost=0; mintime[sx][sy]=0; int min=BFS(start); if(min<Inf) printf("%d/n",min); else printf("Poor ANGEL has to stay in the prison all his life./n"); } return 0;}