Rescue宽度搜索的最优解

Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M &lt;= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.<br><br>Angel's friends want to save Angel. Their task is: approach Angel. We assume that &quot;approach Angel&quot; 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.<br><br>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.)<br>

 

Input

First line contains two integers stand for N and M.<br><br>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. <br><br>Process to the end of the file.<br>

 

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." <br>

 

Sample Input

   
   
   
   

    
    
    
    7 8<br>#.#####.<br>#.a#..r.<br>#..#x...<br>..#..#.#<br>#...##..<br>.#......<br>........<br>
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    13<br>
   
   
   
   

题意：X代表卫兵，a代表终点，r代表起始点，.代表路，#代表墙

路花费一秒，x花费两秒

问到达终点的最少时间

#include<iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;


struct node
{
    int x,y,step;
    friend bool operator<(node n1,node n2)//重载小于号，时间作为比较对象
    {
        return n2.step<n1.step;
    }
};


int n,m,vis[205][205];  //标记
char map[205][205];//地图
int x1,x2,y1,y2;
int to[4][2] = {1,0,-1,0,0,1,0,-1}; //方向数组
int check(int x,int y)
{
    if(x<0 || y<0 || x>=n || y>=m || !vis[x][y] || map[x][y] == '#')
        return 1;
    return 0;
}


int bfs()
{
    int i;
    priority_queue<node> Q;  //优先队列
    node a,next;
    a.x = x1;
    a.y = y1;
    a.step = 0;
    Q.push(a);
    vis[x1][y1] = 0;
    while(!Q.empty())
    {
        a = Q.top();
        Q.pop();
        if(a.x == x2 && a.y == y2)
            return a.step;
        for(i = 0; i<4; i++)
        {
            next = a;
            next.x+=to[i][0];  //方向寻找
            next.y+=to[i][1];
            if(check(next.x,next.y))//判断是否符合要求
                continue;
            next.step++;
            if(map[next.x][next.y] == 'x')//卫兵处多花费了一秒
                next.step++;
            if(vis[next.x][next.y]>=next.step)//存入最小时间
            {
                vis[next.x][next.y] = next.step;
                Q.push(next);  //符合要求就检查下一个
            }
        }
    }
    return 0;
}


int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        for(i = 0; i<n; i++)
        {
            scanf("%s",map[i]);
            for(j = 0; map[i][j]; j++)
            {
                if(map[i][j] == 'r')
                {
                    x1 = i;
                    y1 = j;
                }
                else if(map[i][j] == 'a')
                {
                    x2 = i;
                    y2 = j;
                }
            }
        }
        memset(vis,1,sizeof(vis));
        int ans = 0;
        ans = bfs();
        if(ans)
            printf("%d\n",ans);
        else
            printf("Poor ANGEL has to stay in the prison all his life.\n");
    }


    return 0;
}