ZOJ 649 Rescue(优先队列+bfs)

Rescue

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

Total Submission(s): 521    Accepted Submission(s): 217

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

题意：要求天使的朋友在最短的时间内救出天使，a代表天使，r代表天使的朋友，.代表路，#代表墙，x代表小兵，

每走一步需要一个时间，遇到小兵杀了他要一个时间，移到小兵的位置有要一个时间；

输出最短的时间，

如果救不出来就输出“Poor ANGEL has to stay in the prison all his life.”

 

/*
题解：利用优先队列+bfs
这题一直错，原因竟然是。。。读入有多组数据，我只读了一组
*/
#include 
#include
#include
#include
#include
using namespace std;
int f[202][202];
char mp[202][202];
int n,m,sx,sy;
struct node
{
    int x,y;
    node(int a,int b){x=a; y=b;}
};
int dr[4][2]={{1,0},{0,1},{-1,0},{0,-1} };
struct cmp
{
    bool operator()(node a,node b)
    {
        return f[a.x][a.y]>f[b.x][b.y];
    }
};
int  bfs()
{
    priority_queue,cmp> Q;
    memset(f,0,sizeof(f));
    Q.push(node(sx,sy));
    f[sx][sy]=1;
    while(!Q.empty())
    {
        node p=Q.top();
        Q.pop();
        for(int i=0;i<4;i++)
        {
            int xx=p.x+dr[i][0];
            int yy=p.y+dr[i][1];
            if (xx>=n || xx<0 || yy>=m || yy<0) continue;
            if (mp[xx][yy]=='#' || f[xx][yy]>0) continue;
            f[xx][yy]=f[p.x][p.y]+1;
            if (mp[xx][yy]=='x')   f[xx][yy]++;
            Q.push(node(xx,yy));
            if(mp[xx][yy]=='r') return f[xx][yy]-1;
        }
    }
    return -1;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i)
        {
            scanf("%s",&mp[i]);
            for(int j=0;j)
             if (mp[i][j]=='a') sx=i,sy=j;
        }
        int ans=bfs();
          if (ans==-1) printf("Poor ANGEL has to stay in the prison all his life.\n");
           else printf("%d\n",ans);
    }

    return 0;
}

 

转载于:https://www.cnblogs.com/stepping/p/5764869.html