zoukankan      html  css  js  c++  java
  • POJ 2387 Til the Cows Come Home

    Description:

    Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. 

    Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it. 

    Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

    Input:

    * Line 1: Two integers: T and N 

    * Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

    Output:

    * Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

    Sample Input:

    5 5
    1 2 20
    2 3 30
    3 4 20
    4 5 20
    1 5 100

    Sample Output:

    90

    Hint:

    INPUT DETAILS: 

    There are five landmarks. 

    OUTPUT DETAILS: 

    Bessie can get home by following trails 4, 3, 2, and 1.
     
    题意:有n个点,m条边,现在问奶牛从第n个点到第一个点的最短路径是多少,也就是1~n的最短路径,(毕竟路是双向的,这里发现,POJ好喜欢奶牛的啊~~),直接套用模板中的任何一个就行了。
    #include<stdio.h>
    #include<queue>
    #include<algorithm>
    using namespace std;
    
    const int N=1010;
    const int INF=0x3f3f3f3f;
    
    int G[N][N], dist[N], vis[N], n;
    
    void Init()
    {
        int i, j;
    
        for (i = 1; i <= n; i++)
        {
            dist[i] = INF;
            vis[i] = 0;
            for (j = 1; j <= n; j++)
                G[i][j] = INF;
            G[i][i] = 0;
        }
    }
    
    void Spfa(int u)
    {
        int i, v;
        queue<int>Q;
    
        Q.push(u);
        dist[u] = 0;
    
        while (!Q.empty())
        {
            v = Q.front(); Q.pop();
    
            for (i = 1; i <= n; i++)
            {
                if (dist[i] > dist[v] + G[v][i])
                {
                    dist[i] = dist[v]+G[v][i];
    
                    if (!vis[i])
                    {
                        Q.push(i);
                        vis[i] = 1;
                    }
                }
            }
    
            vis[v] = 0;
        }
    }
    
    int main ()
    {
        int m, a, b, c;
    
        while (scanf("%d%d", &m, &n) != EOF)
        {
            Init();
    
            while (m--)
            {
                scanf("%d%d%d", &a, &b, &c);
    
                G[a][b] = min(G[a][b], c);
                G[b][a] = G[a][b];
            }
    
            Spfa(1);
    
            printf("%d
    ", dist[n]);
        }
    
        return 0;
    }
  • 相关阅读:
    带返回值的多线程
    ORA-12516 "TNS监听程序找不到符合协议堆栈要求的可用处理程序" 解决方案
    Java后端WebSocket的Tomcat实现
    Jackson将对象转换为json字符串时,设置默认的时间格式
    java构造器和构建器
    java静态工厂
    离散-理解只有 才
    数据结构之链式表
    数据结构之顺序线性表
    期末作业验收
  • 原文地址:https://www.cnblogs.com/syhandll/p/4812448.html
Copyright © 2011-2022 走看看