zoukankan      html  css  js  c++  java
  • (广搜) Find a way -- hdu -- 2612

    链接:

    http://acm.hdu.edu.cn/showproblem.php?pid=2612

    Find a way

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 6718    Accepted Submission(s): 2236


    Problem Description
    Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
    Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest. 
    Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.
     
    Input
    The input contains multiple test cases.
    Each test case include, first two integers n, m. (2<=n,m<=200). 
    Next n lines, each line included m character.
    ‘Y’ express yifenfei initial position.
    ‘M’    express Merceki initial position.
    ‘#’ forbid road;
    ‘.’ Road.
    ‘@’ KCF
     
    Output
    For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.
     
    Sample Input
    4 4
    Y.#@
    ....
    .#..
    @..M
    4 4
    Y.#@
    ....
    .#..
    @#.M
    5 5
    Y..@.
    .#...
    .#...
    @..M.
    #...#
     
    Sample Output
    66 88 66

    代码:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <algorithm>
    #include <queue>
    
    using namespace std;
    
    #define INF 0x3f3f3f3f
    #define N 210
    
    struct node
    {
        int x, y, step;
    };
    
    int n, m, a[N][N], dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
    char G[N][N];
    int vis[N][N];
    
    void BFS(node s, int num)
    {
        node p, q;
    
        queue<node>Q;
        Q.push(s);
        memset(vis, 0, sizeof(vis));
        vis[s.x][s.y] = 1;
    
        while(Q.size())
        {
            p = Q.front(), Q.pop();
    
            if(G[p.x][p.y]=='@')
            {
                if(num==1)
                    a[p.x][p.y] = p.step;
                if(num==2)
                    a[p.x][p.y] += p.step;
            }
    
            for(int i=0; i<4; i++)
            {
                q.x = p.x + dir[i][0];
                q.y = p.y + dir[i][1];
                q.step = p.step + 1;
    
                if(!vis[q.x][q.y] && q.x>=0 && q.x<n && q.y>=0 && q.y<m && G[q.x][q.y]!='#')
                {
    
                        vis[q.x][q.y] = 1;
                        Q.push(q);
                }
            }
        }
    }
    
    int main()
    {
        while(scanf("%d%d", &n, &m)!=EOF)
        {
            int i, j;
            node Y, M;
    
            memset(a, 0, sizeof(a));
    
            for(i=0; i<n; i++)
            {
                scanf("%s", G[i]);
                for(j=0; j<m; j++)
                {
                    if(G[i][j]=='Y')
                        Y.x=i, Y.y=j, Y.step=0;
                    if(G[i][j]=='M')
                        M.x=i, M.y=j, M.step=0;
                }
            }
    
            BFS(Y, 1);
            BFS(M, 2);
    
            int ans=INF;
    
            for(i=0; i<n; i++)
            for(j=0; j<m; j++)
                if(G[i][j]=='@' && vis[i][j])
            ans = min(ans, a[i][j]);
    
            printf("%d
    ", ans*11);
    
        }
        return 0;
    }
    勿忘初心
  • 相关阅读:
    整型表示
    有一头母牛,它每年年初生一头小母牛。每头小母牛从第四个年头开始,每年年初也生一头小母牛。请编程实现在第n年的时候,共有多少头母牛?
    shell排序算法实现
    一些小细节
    wordpress新注册用户或重置密码链接失效
    wordpress访问速度慢
    mysql主从复制
    mysql root密码忘记
    mysql root用户登录后无法查看数据库全部表
    Ansible
  • 原文地址:https://www.cnblogs.com/YY56/p/4796657.html
Copyright © 2011-2022 走看看