poj 3268 Silver Cow Party (最短路算法的变换使用【有向图的最短路应用】 )

最新推荐文章于 2021-07-12 19:35:55 发布

转载最新推荐文章于 2021-07-12 19:35:55 发布 · 84 阅读

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原文链接：http://www.cnblogs.com/yspworld/p/4372506.html

本文探讨了SilverCowParty问题，即每头牛从各自农场出发到派对再返回的最短时间路线选择问题。该问题通过使用Dijkstra算法进行解决，并提供了完整的代码实现。代码中包括了两轮Dijkstra算法的应用，分别用于计算到派对地点和返回的最短时间。

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Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13611		Accepted: 6138

Description

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T_i (1 ≤ T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2..M+1: Line i+1 describes road i with three space-separated integers: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

Sample Output

10

题目及算法分析：
代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <iostream>
#include <string>
#include <stack>
#include <queue>
#include <algorithm>
#define N 1000+20
#define INF 0x3f3f3f3f

using namespace std;

int map[N][N];
int dis[N], ans[N];
bool vis[N];
int n, m, s;

int Dijkstra(int s)
{
	int i, j, k;
	for(i=1; i<=n; i++)
		dis[i]=map[s][i];
	vis[s]=true;
	for(k=1; k<n; k++)
	{
		int mi=INF, pos;
		for(i=1; i<=n; i++)
		{
			if(vis[i]==false && dis[i]<mi )
			{
				mi=dis[i]; pos=i;
			}
		}
		vis[pos]=true;
		for(j=1; j<=n; j++)
		{
			if(vis[j]==false && dis[j]>dis[pos]+map[pos][j] )
				dis[j]=dis[pos]+map[pos][j];
		}
	}
	for(i=1; i<=n; i++)
	{
		ans[i]=ans[i]+dis[i];
	}
	return 0;
}

void turn_over()
{
	for(int i=1; i<=n; i++)
	{
		for(int j=1; j<i; j++)
			swap(map[i][j], map[j][i] );
	}
}

int main()
{
	scanf("%d %d %d", &n, &m, &s);
	int u, v, w;

	for(int i=1; i<=n; i++)
		for(int j=1; j<=n; j++)
		{
			if(i==j) map[i][j]=0;
			else map[i][j]=INF;
		}

	for(int i=0; i<m; i++)
	{
		scanf("%d %d %d", &u, &v, &w);
		map[u][v] = w;
	}
	memset(vis, false, sizeof(vis));
	memset(ans, 0, sizeof(ans));
	Dijkstra(s);
	turn_over();
	memset(vis, false, sizeof(vis));
    Dijkstra(s);
	int cc=-1;
    for(int i=1; i<=n; i++)
    {
        if(ans[i]>cc && ans[i]<INF )
            cc=ans[i];
    }
    printf("%d\n", cc );
	return 0;
}

转载于:https://www.cnblogs.com/yspworld/p/4372506.html