Silver Cow Party

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; roadi 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.

Source

USACO 2007 February Silver

题目的意思是：从出发点到达目的地X，再从X返回到出发点的最短路径中的最大值（因为出发点没有固定，也就是可以：1->X->1, 2->X->2等）。所以我们要枚举所有的可能性，找出其中的最大值。巧妙地运用dijkstra算法，双向求出两次X->m的最短路径长然后相加即得到了m->X->m的最短路径。代码如下：

#include<queue>
#include<vector>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct Edge
{
    int to;
    int dis;
    Edge(int to, int dis){
        this -> to = to;
        this -> dis = dis;
    }
};
typedef pair<int,int>P;

int a,b,c;
int N,M,X;
int d1[1005],d2[1005];
vector<Edge> G1[1005];
vector<Edge> G2[1005];
void dijkstra(int s,int d[],vector<Edge> G[])
{
    priority_queue<P,vector<P>,greater<P> >q;
    d[s]=0;
    q.push(P(0,s));
    while(q.size())
    {
        P p=q.top();
        q.pop();
        int v=p.second;
        for(int i=0;i<G[v].size();i++)
        {
            Edge& e=G[v][i];
            if(d[e.to]>d[v]+e.dis)
            {
                d[e.to]=d[v]+e.dis;
                q.push(P(d[e.to],e.to));
            }
        }
    }
}
int main()
{
    memset(d1,0x5f,sizeof(d1));
    memset(d2,0x5f,sizeof(d2));
    scanf("%d%d%d",&N,&M,&X);
    for(int i=1;i<=M;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        G1[a].push_back(Edge(b,c));
        G2[b].push_back(Edge(a,c));
    }
    dijkstra(X,d1,G1);
    dijkstra(X,d2,G2);
    int small_max=-1;
    for(int i=1;i<=N;i++)
    {
        if(i==X) continue;
        small_max=max(small_max,d1[i]+d2[i]);
    }
    cout<<small_max<<endl;
}