[kuangbin带你飞]专题四最短路练习 D - Silver Cow Party

最新推荐文章于 2021-05-04 13:37:00 发布

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本文详细解析了POJ-3268问题中寻找从各农场到聚会地点并返回的最长往返时间的算法。通过使用Dijkstra算法两次分别计算到聚会地点和从聚会地点返回的最短路径，最后求得所有路径中的最大值。文章提供了完整的C++代码实现。

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POJ - 3268

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 Ti (1 ≤ Ti ≤ 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: Ai,Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.

Output

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

Sample Input

Sample Output

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

#include <iostream>
#include <cstring>
#include <algorithm>
#define inf 99999999
using namespace std;

int n,m,x;
int map[1010][1010];
int vis[1010],d[1010],p[1010];


void dij()
{
	int v;
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++)
	{
		d[i]=map[x][i];
		p[i]=map[i][x];
	}
	for(int i=1;i<=n;i++)//过去 
	{  
        int minn=inf;
        for(int j=1;j<=n;j++)  
            if(!vis[j]&&d[j]<minn)
			{
                v=j;  
                minn=d[j];
            } 
        vis[v]=1;
        for(int j=1;j<=n;j++)  
            if(!vis[j]&&d[j]>map[v][j]+d[v])  
			    d[j]=map[v][j]+d[v];  
    }
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)//回来 
	{  
        int minn=inf;
        for(int j=1;j<=n;j++)  
            if(!vis[j]&&p[j]<minn)
			{
                v=j;
                minn=p[j];
            } 
        vis[v]=1;
        for(int j=1;j<=n;j++)  
            if(!vis[j]&&p[j]>map[j][v]+p[v])  
			    p[j]=map[j][v]+p[v]; 
    }
}

int main()
{
	scanf("%d%d%d",&n,&m,&x);
	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++)
	{
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		map[a][b]=c;
	}
	dij();
	for(int i=1;i<=n;i++)d[i]+=p[i];
	sort(d+1,d+1+n);
	cout<<d[n]<<endl;
	return 0;
}