hdu 1242 Rescue bfs+优先队列priority

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我是靠谱客的博主调皮烤鸡，最近开发中收集的这篇文章主要介绍hdu 1242 Rescue bfs+优先队列priority，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Rescue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18318 Accepted Submission(s): 6544

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

题意: 求r到a最短距离,遇到守卫x ,要多花一秒.

因为有守卫,所以普通队列得到的不一定是最优解,普通队列中最前面的只是步数最小,不是时间最少.所以要用优先队列,让时间花费最小的排在前面.

注意: 因为没有考虑到可能没有r,也就是没人来救的情况,结果我一直RE. 衰.

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

int dir[4][2]={1,0,-1,0,0,1,0,-1};
char mp[2100][2100];
int vis[2100][2100];
struct nodes
{
    int x,y;
    int t;
    friend bool operator <(const nodes&a,const nodes&b)
    {
        return a.t>b.t;
    }
}; 
int kk;
int bfs(int n,int m)
{ 
    priority_queue<nodes>q;
    nodes now,tem; 
    memset(vis,0,sizeof vis); 
	int flag=0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(mp[i][j]=='r')
            {
                vis[i][j]=1;
                now.x=i;
                now.y=j;
                now.t=0;
				flag=1;
            } 
        }
    }
    if(flag==0)
		return -1; 
    int xx,yy; 
    q.push(now);
    while(!q.empty())
    { 
        now=q.top();
        q.pop(); 
      
        if(mp[now.x][now.y]=='a')
            return now.t; 
        for(int i=0;i<4;i++)
        { 
            xx=now.x+dir[i][0];
            yy=now.y+dir[i][1];
            if(xx>=0&&xx<n&&yy>=0&&yy<m)
            {
                if(vis[xx][yy]==0&&mp[xx][yy]!='#')
                {
                    
                    
                    tem.x=xx;
                    tem.y=yy;
                    if(mp[xx][yy]=='.'||mp[xx][yy]=='a') 
                        tem.t=now.t+1; 
                    else if(mp[xx][yy]=='x')
                        tem.t=now.t+2; 
                    
                    vis[xx][yy]=1;
                    q.push(tem);
                }
            }
        }  
    } 
    return -1; 
}


int main()
{ 
    int n,m; 
    while(scanf("%d%d",&n,&m)!=EOF)
    { 
        for(int i=0;i<n;i++)
            scanf("%s",mp[i]); 
        int ans=bfs(n,m); 
        if(ans==-1) 
            puts("Poor ANGEL has to stay in the prison all his life."); 
        else
            printf("%dn",ans); 
    } 
    return 0;
}