POJ2387 Til the Cows Come Home 最短路

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本文介绍了一个经典的最短路径问题，通过Dijkstra算法解决Bessie从田野返回牛棚的最短路径问题。文章详细展示了算法实现，包括初始化、松弛操作和更新最短路径的过程。

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Til the Cows Come Home

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 91593 Accepted: 29824
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

问题连接

问题描述

有N个顶点，T条边，T行说明每条边的长度和关联的顶点。找出从顶点n到顶点1的最短路径长。

问题分析

最短路问题。用dijkstra算法做，算是模板题吧。注意题目有点坑，图有重边，构图的时候，选择最短的边。

代码如下

#include<iostream>
using namespace std;

const int N = 1e3 + 5;
const int inf = 0x3f3f3f3f;

int n;
int G[N][N];
int dist[N];
bool vis[N];

void init() {//初始化
	for (int i = 1; i <= n; i++)
		for (int j = i; j <= n; j++)
			if (i == j) G[i][i] = 0;
			else G[i][j] = G[j][i] = inf;
}

void dijkstra(int start) {
	int i, j, minn;
	for (i = 1; i <= n; i++) {
		dist[i] = G[start][i];
		vis[i] = 0;
	}
	dist[start] = 0;
	vis[start] = 1;
	for (i = 1; i < n; i++) {
		minn = inf;
		for (j = 1; j <= n; j++) {
			if (minn > dist[j]&&!vis[j]) {
				minn = dist[j];
				start = j;
			}
		}
		if (minn == inf) break;
		vis[start] = 1;
		for (j = 1; j <= n; j++)
			if (!vis[j] && dist[j] > dist[start] + G[start][j]) dist[j] = dist[start] + G[start][j];
	}
}


int main() {
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	int t, i, v, u, e;
	cin >> t >> n;
	init();
	while (t--) {
		cin >> v >> u >> e;
		if(e<G[v][u]) G[v][u] = G[u][v] = e;//处理重边
	}
	dijkstra(n);
	cout << dist[1] << endl;
	return 0;
}