HDU-1242 Rescue

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

Sample Output
13

分析：使用全局队列，多点同时bfs求出最短路，若单独求会超时。

Source：

#include<cstdio>#include<iostream>#include<queue>#include<cstring>#include<algorithm>using namespace std;const int maxn=1005;const int INF=0x3f3f3f3f;int N,M;char map[maxn][maxn];bool vis[maxn][maxn];int dx[4]={0,1,0,-1};int dy[4]={1,0,-1,0};int path[maxn][maxn];   //记录步数queue<int> qx;queue<int> qy;   //全局队列int bfs(){    while(!qx.empty()&&!qy.empty())    {        int x=qx.front();        int y=qy.front();        qx.pop();        qy.pop();        for(int i=0;i<4;i++)        {            int x1=x+dx[i];            int y1=y+dy[i];            if(x1>=0&&x1<N&&y1>=0&&y1<M&&!vis[x1][y1]&&map[x1][y1]!='#')            {                qx.push(x1);                qy.push(y1);                vis[x1][y1]=true;                path[x1][y1]=path[x][y]+1;                if(map[x1][y1]=='x')                    path[x1][y1]++;                if(map[x1][y1]=='a')                    return path[x1][y1];            }        }    }    return INF;}int main(){    while(scanf("%d%d",&N,&M)!=EOF)    {        for(int i=0;i<N;i++)            for(int j=0;j<M;j++)                cin>>map[i][j];        memset(vis,false,sizeof(vis));        memset(path,-1,sizeof(path));        while(qx.size())            qx.pop();        while(qy.size())            qy.pop();   //初始化，清空队列        for(int i=0;i<N;i++)        {            for(int j=0;j<M;j++)            {                if(map[i][j]=='r')                {                    qx.push(i);                    qy.push(j);   //多点入队                    vis[i][j]=true;                    path[i][j]=0;   //初始化                }            }        }        int k=bfs();        if(k<INF)            cout<<k<<endl;        else            cout<<"Poor ANGEL has to stay in the prison all his life."<<endl;    }    return 0;}