POJ 2387 Til the Cows Come Home (dijkstra)-优快云博客

本文介绍了一种基本的最短路径算法——Dijkstra算法，并通过实例演示了如何解决从任意点到起始点的最短路径问题。通过解析样例输入和输出，读者可以理解算法的核心思想和实现细节。

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最简单的最短路，dijkstra入门水题

Til the Cows Come Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 35217		Accepted: 11946

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

Sample Output

Hint

INPUT DETAILS:

There are five landmarks.

OUTPUT DETAILS:

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

代码

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<math.h>
#include<string>
#include<set>
#include<queue>
#include<map>
using namespace std;
const int INF=999999999;
int n,m;
int G[1110][1110];
bool vis[1111];
int d[1111];
int dijkstra(int st,int ed)
{
    memset(vis,false,sizeof(vis));
    int i,j;
    for(i=1;i<=n;i++)
        d[i]=G[st][i];
    d[st]=0;
    vis[st]=true;
    for(i=2;i<=n;i++)
    {
        int w=0,minn=INF;
        for(j=1;j<=n;j++)
        {
            if(vis[j]==false&&d[j]<minn)
            {
                minn=d[j];
                w=j;
            }
        }//找到没访问过的当前最近的点w
        vis[w]=true;//访问w
        for(int k=1;k<=n;k++)//更新w
        {
            if(vis[k]==false&&G[k][w]!=INF&&d[k]>G[w][k]+d[w])
            {
                d[k]=G[w][k]+d[w];
            }
        }
    }
    return d[ed];
}
int main()
{
//    freopen("D://input.txt","r",stdin);
    while(scanf("%d%d",&m,&n)!=EOF)//n表示点，m表示路
    {
        int i,j;
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                    G[i][j]=INF;
        while(m--)
        {
            int s,e,l;
            scanf("%d%d%d",&s,&e,&l);
            G[s][e]=min(l,G[s][e]);
            G[e][s]=G[s][e];
        }
        int ans=dijkstra(1,n);
        printf("%d\n",ans);
    }
    return 0;
}