简单广搜：Rescue （优先队列）

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.) 

InputFirst 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. 
OutputFor 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

思路：这道题样例十分水，因为不用优先队列也可以过这组样例，所以为了避免其他人犯同样的错误，本博主就再弄一组样例：

输入：

7 8
a######r
........
#..#x...
..#..#.#
#...##..
.#......
........

输出：9

正确代码：

#include
#include
#include
#include
#include
#include
#include
using namespace std;
char map[210][210];
int d[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};
int n,m;
struct node
{
    int x,y,t;
} q,p;
struct cmp{
    bool operator () (node &a,node &b)
    {
        return a.t>b.t;
    }

};//优先队列模板。
bool check(int p,int pp)
{
    if(p<0||pp<0||p>=m||pp>=n||map[p][pp]=='#')
        return 0;
    return 1;
}
void bfs()
{
    priority_queue,cmp>que;//创建一个优先队列。最小值优先。
    que.push(q);
    while(!que.empty())
    {
        q=que.top();
        que.pop();
        for(int i=0; i<4; i++)
        {
            p.x=q.x+d[i][0];
            p.y=q.y+d[i][1];
            if(check(p.x,p.y))
            {
                if(map[p.x][p.y]=='x')
                    p.t=q.t+2;
                else
                    p.t=q.t+1;
                if(map[p.x][p.y]=='a')
                {
                    printf("%d\n",p.t);
                    return ;
                }
                que.push(p);
                map[p.x][p.y]='#';
            }
        }
    }
    printf("Poor ANGEL has to stay in the prison all his life.\n");
}
int main()
{
    while(~scanf("%d%d%*c",&m,&n))
    {
        memset(map,0x00,sizeof(map));
        for(int i=0; i