BFS专题之hdu1242 rescue

http://acm.hdu.edu.cn/showproblem.php?pid=1242

有点简单的  BFS题目  唯一的不同 是X点可以 停滞一下

这次采取的方法与以前不同  对于X点的处理

注意 VIS【】【】记录是否访问和 走过此点的最短时间

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
char map[205][205];
int vis[205][205];
struct node
{
    int x,y;
    int s;
}dui[80005];
int dir[4][2]={0,1,0,-1,1,0,-1,0};
int n,m,s_x,s_y;
void bfs()
{
    int i,j;
    int head,tail;
    struct node next,now;
    memset(vis,0,sizeof(vis));
    head=tail=0;
    dui[tail].x=s_x;
    dui[tail].y=s_y;
    dui[tail++].s=1;
    vis[s_x][s_y]=1;
    while(head<tail)
    {
        now.x=dui[head].x;
        now.y=dui[head].y;
        now.s=dui[head++].s;
        if(map[now.x][now.y]=='x')
        {
            map[now.x][now.y]='.';
            dui[tail].x=now.x;
            dui[tail].y=now.y;
            dui[tail++].s=now.s+1;
            continue;
        }
        for(i=0;i<4;i++)
        {
            next.x=now.x+dir[i][0];
            next.y=now.y+dir[i][1];
            next.s=now.s+1;
            if( next.x<0 || next.y<0 || next.x>=n || next.y>=m || map[next.x][next.y]=='#')
                continue;
            if(vis[next.x][next.y]!=0) continue;
            vis[next.x][next.y]=next.s;
            dui[tail].x=next.x;
            dui[tail].y=next.y;
            dui[tail++].s=next.s;
        }
    }
}
int main()
{
    int i,j,ans;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        ans=10000000;
        getchar();
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                scanf("%c",&map[i][j]);
                if(map[i][j]=='a')
                {
                    s_x=i; s_y=j;
                }
            }
            getchar();
        }
        bfs();
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(map[i][j]=='r')
                {
                    if(vis[i][j]==0) continue;
                    else
                    {
                        if(ans>vis[i][j]) ans=vis[i][j];
                    }
                }
            }
        }
        if(ans==10000000) printf("Poor ANGEL has to stay in the prison all his life.\n");
        else printf("%d\n",ans-1);
    }
    return 0;
}