D

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.

 

 

题意：

有编号为1－N的牛，它们之间存在一些单向的路径。给定一头牛的编号，其他牛要去拜访它并且拜访完之后要返回自己原来的位置，求这些牛中所花的最长的来回时间是多少。

思路：

直接以被拜访的牛作为起点，让他去拜访其他牛。就是单源最短路径。但是这题很巧妙的地方在于如何转化路径的方向，而不用多花费时间与空间

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <stdbool.h>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <math.h>
#include <queue>
#include <stack>
#include <map>

#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int MAXN = 1005;

int T,n,m;
int graph[MAXN][MAXN];
int dist[MAXN];
bool vis[MAXN];
int ans[MAXN];
int x;

void dijistra(int x)
{
    int pos = x;
    for (int i=1;i<=n;i++)
        dist[i] = graph[pos][i];
    vis[pos] = true;
    for (int i=1;i<=n;i++)
    {
        int mins = INF;
        for (int j=1;j<=n;j++)
        {
            if (!vis[j] && dist[j]<mins)
            {
                pos = j;
                mins = dist[j];
            }
        }
        vis[pos] = true;
        for (int j=1;j<=n;j++)
        {
            if (!vis[j] && dist[j] > dist[pos]+graph[pos][j])
                dist[j] = dist[pos]+graph[pos][j];
        }
    }
}

void change()  //这个地方非常巧妙，转变方向
{
    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<=i;j++)
            swap(graph[i][j],graph[j][i]);
    }
}

int main()
{
    //freopen("../in.txt","r",stdin);
    while (~scanf("%d%d%d",&n,&m,&x)) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                if (i == j)
                    graph[i][j] = 0;
                else
                    graph[i][j] = INF;
            }
        }
        for (int i=1;i<=m;i++){
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            graph[a][b] = c;
        }
        dijistra(x);
        for (int i=1;i<=n;i++)
            ans[i] = dist[i];
        change();
        memset(vis,false, sizeof(vis));
        dijistra(x);
        int cnt = 0;
        for (int i=1;i<=n;i++)
        {
            ans[i] += dist[i];
            cnt = max(cnt,ans[i]);
        }
        printf("%d
",cnt);
    }
    return 0;
}