poj 3268 Silver Cow Party ([kuangbin带你飞]专题四 最短路练习)

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

4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3

Sample Output

10

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.

题目大意：给出n个点和m条边，接着是m条边，代表从牛a到牛b需要花费c时间，现在所有牛要到牛x那里去参加聚会，并且所有牛参加聚会后还要回来，给你牛x，除了牛x之外的牛，他们都有一个参加聚会并且回来的最短时间，从这些最短时间里找出一个最大值输出

解题思路：求牛X到其余牛的最短路径的2倍就可以了，单元最短路

#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 1000+10;
int n,m,x;
struct Node{
	int to;int cost;
	Node(){};
	Node(int x,int y):to(x),cost(y){};
}node;
vector<Node> E[maxn];
typedef pair<int,int> P;
int d[maxn][maxn];
void dij(int s){
	//
	priority_queue<P,vector<P>,greater<P> > pq;
	memset(d[s],0x3f,sizeof(d[s]));
	d[s][s]=0;
	pq.push(P(0,s));
	while(!pq.empty()){
		P p= pq.top();
		pq.pop();
		int v=p.second;
	//	if(d[s][v] < p.first) continue;
		for(int i=0;i<E[v].size();i++){
			int u=E[v][i].to;
			if(d[s][u] > d[s][v]+E[v][i].cost){
				d[s][u] = d[s][v] + E[v][i].cost;
				pq.push(P(d[s][u],u));
			}
		}
	}

}




int main(){
	scanf("%d%d%d",&n,&m,&x);
	for(int i=0;i<m;i++){
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		E[x].push_back(Node(y,z));
		//E[y].push_back(Node(x,z));
	}
	for(int i=1;i<=n;i++)
		dij(i);
	int ans=0;
	for(int i=1;i<=n;i++){
		if(i==x){
			continue;
		}
		ans=max(ans,d[i][x]+d[x][i]);
	}
	printf("%d\n",ans);

	return 0;
}