hdu 1242 Rescue

问题描述

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.)

 

输入

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.

 

输出

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."

 

样例输入

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

 

样例输出

 13 

这题与poj坦克大战那道题类似http://blog.csdn.net/qq_31237061/article/details/50593233

遇到卫兵多花1秒时间；

思路：bfs+优先队列：

代码：

#include <iostream>#include <queue>#include <cstdio>#include <cstring>using namespace std;struct node{    int x,y,step;    node(int a,int b,int c)    {        x=a;y=b;step=c;    }    bool operator < (const node & a)const    {        return step> a.step;    }};const int maxx=205;priority_queue <node> que;char ma[maxx][maxx];int vis[maxx][maxx];int n,m;int xx,yy,ex,ey;int dx[]={1,0,-1,0};int dy[]={0,1,0,-1};int bfs(){    node no(xx,yy,0);    vis[xx][yy]=1;    que.push(no);    while(!que.empty())    {        node no1=que.top();        que.pop();        for(int i=0;i<4;i++)        {            int x=no1.x+dx[i];            int y=no1.y+dy[i];            int step=no1.step;            if(vis[x][y]||x<0||x>=n||y<0||y>=m||ma[x][y]=='#')continue;            if(ma[x][y]=='x')                step++;            if(ma[x][y]=='a')                return step+1;            vis[x][y]=1;            que.push(node(x,y,step+1));        }    }    return -1;}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        while(!que.empty())que.pop();        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            scanf("%s",&ma[i]);            for(int j=0;j<m;j++)            {                if(ma[i][j]=='r')                {                    xx=i;yy=j;                }            }        }        int c;        c=bfs();        if(c!=-1)            printf("%d\n",c);        else            printf("Poor ANGEL has to stay in the prison all his life.\n");    }}