POJ 2387 Til the Cows Come Home

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原文链接：http://www.cnblogs.com/Cantredo/p/9880275.html

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#数据结构与算法

在一片广阔的牧场中，Bessie需要找到从苹果树林回到谷仓的最短路径，以确保她能获得足够的睡眠。这是一道典型的图论问题，通过使用Dijkstra算法，我们可以在一系列地标和路径中找到最短的距离。

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

Hint

INPUT DETAILS:

There are five landmarks.

OUTPUT DETAILS:

Bessie can get home by following trails 4, 3, 2, and 1.

Source

USACO 2004 November

题解

dijkstra寻找最短路模版题。代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std ;

const int INF = 0x3f3f3f3f ;
const int MAXN = 2005 ;
int dis[MAXN] ;
int map[MAXN][MAXN] ;
bool vis[MAXN] ;
int n , m ;

void dijkstra(){
    int cur = 1 ;
    dis[cur] = 0 ;
    vis[cur] = true ;
    for ( int i = 1 ; i <= n ; i ++ ){
        for ( int j = 1 ; j <= n ; j ++ ){
            if ( !vis[j] && map[cur][j] + dis[cur] < dis[j] ){
                dis[j] = map[cur][j] + dis[cur] ;
            }
        }
        int mini = INF ;
        for ( int j = 1 ; j <= n ; j ++ ){
            if ( !vis[j] && dis[j] < mini ){
                mini = dis[cur=j] ;
            }
        }
        vis[cur] = true ;
    }
}

void init(){
    memset(vis , false , sizeof(vis)) ;
    for ( int i = 1 ; i <= n ; i ++ ){
        dis[i] = INF ;
        for ( int j = 1 ; j <= n ; j ++ ){
            map[i][j] = INF ;
        }
    }
    return ;
}

int main(){
    while ( cin >> m >> n ){
        init() ;
        for ( int i = 0 ; i < m ; i ++ ){
            int from , to , val ;
            cin >> from >> to >> val ;
            if ( map[from][to] > val ){
                map[to][from] = map[from][to] = val ;
            }
        }
        dijkstra() ;
        cout << dis[n] << endl ;
    }
    return 0 ;
}

转载于:https://www.cnblogs.com/Cantredo/p/9880275.html