HDU1242 Rescue BFS Rescue

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我是靠谱客的博主勤奋果汁，最近开发中收集的这篇文章主要介绍HDU1242 Rescue BFS Rescue，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

BFS+优先队列即可。

Rescue

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

Total Submission(s): 10964 Accepted Submission(s): 4018

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>

using namespace std;

bool vis[210][210];
char g[210][210];
int n,m;
int x1,y1,x2,y2;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
struct node
{
    int x,y,t;
    bool operator<(const node &a)const
    {
        return t>a.t;
    }
};

bool jud(int x,int y)
{
    if(x<0||x>=n||y<0||y>=m)
        return false;
    if(g[x][y]=='#'||vis[x][y])
        return false;
    return true;
}

int bfs()
{
    node a,k;
    priority_queue<node> q;
    a.x=x1,a.y=y1,a.t=0;
    vis[x1][y1]=1;
    q.push(a);
    while(!q.empty())
    {
        a=q.top();
        q.pop();
        for(int i=0;i<4;i++)
        {
            k.t=a.t+1;
            k.x=a.x+dir[i][0];
            k.y=a.y+dir[i][1];
            if(!jud(k.x,k.y)) continue;
            if(g[k.x][k.y]=='x') k.t++;
            if(k.x==x2 && k.y==y2)
                return k.t;
            vis[k.x][k.y]=1;
            q.push(k);
        }
    }
    return -1;
}

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