POJ 3268 Silver Cow Party

最新推荐文章于 2021-04-01 18:48:34 发布

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本文介绍了一个典型的最短路径问题，即求解每头奶牛从各自农场到派对地点并返回所需的最短时间，采用Dijkstra算法进行求解。

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Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 23108		Accepted: 10568

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.

题意：有N只奶牛要去X奶牛家参加party，有M条路，他们不但要去，还要回来，求出最短路的最大值

赤裸裸的最短路，难点在于求一个点到多点的最短路，再求多点到一个点的最短路。其实解决方法很简单，转一下数组就好了嘛

用的是dijkstra算法

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int inf=0x3f3f3f3f;
int n,m,x;
int map[1010][1010];
int disFirst[1010];
int disSecond[1010];
int book[1010];
void dijkstra(int *temp){
	memset(book,0,sizeof(book));
	for(int i=1;i<=n;i++) temp[i]=map[x][i];
	book[x]=1;
	for(int i=1;i<=n-1;i++){
		int min=inf,flag;
		for(int j=1;j<=n;j++){
			if(!book[j]&&temp[j]<min){
				min=temp[j];
				flag=j;
			}
		}
		book[flag]=1;
		for(int j=1;j<=n;j++){
			if(!book[j]&&temp[j]>temp[flag]+map[flag][j]){
				temp[j]=temp[flag]+map[flag][j];
			}
		}
	}
}
int main(){
	while(~scanf("%d%d%d",&n,&m,&x)){
		for(int i=0;i<1010;i++){
			for(int j=0;j<1010;j++){
				if(i==j) map[i][j]=0;
				else map[i][j]=inf;
			}
		}
		while(m--){
			int a,b,c;
			cin>>a>>b>>c;
			if(map[a][b]>c) map[a][b]=c;
		}
		dijkstra(disFirst);
		for(int i=1;i<=n;i++){
			for(int j=1;j<=i;j++){
				swap(map[i][j],map[j][i]);
			}
		}
		dijkstra(disSecond);
		int ans=-1;
		for(int i=1;i<=n;i++){
			ans=max(ans,disFirst[i]+disSecond[i]);
		}
		cout<<ans<<endl;
	}
}