1242 Rescue【bfs】

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 21316    Accepted Submission(s): 7598

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

找最短路径，bfs 题型，因为中间过程可能会影响时间的使用不均衡，所以这个题要用优先队列来解决，优先考虑用时间少的方法，把每一个状态拓展来的状态都放进队列里，注意标记重复的状态.....从天使的位置开始考虑比较好，因为 r 可能有多个，只需要找到最近的那个 r 就行！！

#include<stdio.h>#include<string.h>#include<queue>using namespace std;char x[205][205];int v[205][205],n,m,kase;int fx[4]={-1,0,0,1},fy[4]={0,-1,1,0};struct save{    int x,y,t;    friend bool operator<(save a,save b)    {        return a.t>b.t;//时间少的优先考虑    }}r,t;priority_queue<save> q;void bfs(){    while(!q.empty())    {        r=q.top();        q.pop();        if(x[r.x][r.y]=='r')//找到了一个，立刻终止        {            kase=1;            return;        }        for(int i=0;i<4;++i)//向四个方向拓展        {            int tx=fx[i]+r.x,ty=fy[i]+r.y;            t.x=tx; t.y=ty; t.t=r.t+1;            if(x[tx][ty]=='#')            {                continue;            }            if(x[tx][ty]=='x')//有守卫，多费一个时间            {                ++t.t;            }            if(tx>=0&&tx<n && ty>=0&&ty<m&&!v[tx][ty])            {                v[tx][ty]=1;//用过的标记                q.push(t);            }        }    }}int main(){    int i,j,a,b;char y;    while(~scanf("%d%d",&n,&m))    {        for(i=0;i<n;++i)        {            getchar();            for(j=0;j<m;++j)            {                scanf("%c",&y);                x[i][j]=y;                if(y=='a')                {                    a=i;b=j;                }            }        }        while(!q.empty())//注意清空        {            q.pop();        }        memset(v,0,sizeof(v));        kase=0;        r.x=a; r.y=b; r.t=0;        q.push(r);v[a][b]=1;//把状态入队列....        bfs();        if(kase==1)        {            printf("%d\n",r.t);            continue;        }        printf("Poor ANGEL has to stay in the prison all his life.\n");    }    return 0;}