hdu 1242 Rescue

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13510    Accepted Submission(s): 4943

Problem Description

 

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
 
 

#include"stdio.h"#include"queue"using namespace std;#define N 210char map[N][N];int dir[4][2]={1,0,-1,0,0,-1,0,1};int n,m;struct node{int x,y,step;friend bool operator<(node a,node b){return a.step>b.step;     //步数少的优先  }};int bfs(int x,int y){priority_queue<node>q;node cur,next;int i,di,dj;map[x][y]='#';cur.x=x;cur.y=y;cur.step=0;q.push(cur);while(!q.empty()){cur=q.top();          //取步数最少的点先走q.pop();              //删除顶端元素for(i=0;i<4;i++)           //每次从一点走向四个方向{di=cur.x+dir[i][0];dj=cur.y+dir[i][1];if(di>=0&&di<n&&dj>=0&&dj<m&&map[di][dj]!='#'){next.x=di;next.y=dj;                next.step=cur.step+1;if(map[di][dj]=='r')       //找到朋友返回最少步数return next.step;else if(map[di][dj]=='.')q.push(next);else          {next.step++;q.push(next);}map[di][dj]='#';    //不能重复走，否则队列超内存}}}return -1;}int main(){int i,j,ans,si,sj;while(scanf("%d%d",&n,&m)!=-1){for(i=0;i<n;i++){scanf("%s",map[i]);for(j=0;j<m;j++)if(map[i][j]=='a'){si=i;sj=j;}}ans=bfs(si,sj);if(ans==-1)printf("Poor ANGEL has to stay in the prison all his life.\n");elseprintf("%d\n",ans);}return 0;}