POJ - 3268 Silver Cow Party-优快云博客

本文链接：https://blog.youkuaiyun.com/jerans/article/details/61630668

本文介绍了一个算法问题，即如何找到N个农场中某头奶牛参加聚会并返回所需最长总时间的问题。通过建立两个图，并使用两次迪杰斯特拉算法（Dijkstra），分别计算从聚会地点到各农场的最短路径及返回路径，从而高效地解决了该问题。

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Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 20881		Accepted: 9550

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

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.

Source

USACO 2007 February Silver

把边正向建一个图，在反向建一个图，两遍dj，正向建是求回去的最短路径，而反向建是因为用dj求每个点到单点的最短距离复杂度很高，反向建图后求终点到每个点的最短路径就可以了

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define eps 1e-9
#define PI 3.141592653589793
#define bs 1000000007
#define bsize 256
#define MEM(a) memset(a,0,sizeof(a))
#define inf 0x3f3f3f3f
#define rep(i,be,n) for(i=be;i<n;i++)
typedef long long ll;
using namespace std;
int mp[2][1005][1005];
int book[1005];
int x,n,m;
int dijis(int k)
{
	int index,i,j,minn;
	MEM(book);
	book[x]=1;
	for(i=1;i<=n;i++)
	{	
		minn=inf;
		for(j=1;j<=n;j++)
		{
			if(!book[j]&&mp[k][x][j]<minn)
			{
				index=j;
				minn=mp[k][x][j];
			}
		}
		book[index]=1;
		for(j=1;j<=n;j++)
		{
			if(!book[j]&&mp[k][index][j]+minn<mp[k][x][j])
			{
				mp[k][x][j]=mp[k][index][j]+minn;
			}
		}
	}
}
int main()
{
	int i,j,a,b,t;
	memset(mp,inf,sizeof(mp));
	scanf("%d %d %d",&n,&m,&x);
	while(m--)
	{
		scanf("%d %d %d",&a,&b,&t);
		if(mp[0][a][b]>t)
		{
			mp[0][a][b]=t;
		}
		if(mp[1][b][a]>t)
		{
			mp[1][b][a]=t;
		}
	}
	dijis(0);
	dijis(1);
	int ans=0;
	for(i=1;i<=n;i++)
	{
		if(i==x)
		continue;
		ans=max(ans,mp[0][x][i]+mp[1][x][i]);
	}
	printf("%d\n",ans);
	return 0;
 }