POJ - 3268 Silver Cow Party

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原创最新推荐文章于 2026-09-19 14:59:47 发布 220 阅读 AI 写同款· GEO 优化 ›

【代码超详解】POJ 3268 Silver Cow Party（Dijkstra 算法 + 思维）一、题目描述二、算法分析说明与代码编写指导每头牛都会到点 x 参加聚会，然后返回。但路是单向的，因此回去的路可能与来的路不同。如果对每个点都跑一次 Dijkstra，或者直接跑一次 Floyd 算法，由于复杂度达到 O(n^3)，在本题的数据规模下会超时。从 x 回去的最短路可以直接套 Dijkstra 算法求解单源最短路径。从各点来到 x 的最短路径也只需要跑一次 Dijkstra 就... 阅读详情

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;
 }