最短路径（巧妙的矩阵交换）

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 35984		Accepted: 16104

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：去往并返回

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdio.h>
#include <queue>
#include <string.h>
#include <vector>
#include <map>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 1000000007
using namespace std;
typedef long long ll ;
int n , m ;
int ma[1109][1109];
int dis[1109];
int vis[1109] ;

int d[1109];


void Dijia(int r)
{
    for(int i = 1 ; i <= n ; i++)
    {
        vis[i] = 0 ;
        dis[i] = ma[r][i];
    }
    vis[r] = 1 ;
    for(int i = 1 ; i < n ; i++)
    {
        int min1 = INF ;
        int pos ;
        for(int j = 1 ; j <= n ; j++)
        {
            if(!vis[j] && dis[j] < min1)
            {
                min1 = dis[j];
                pos = j ;
            }
        }
        vis[pos] = 1 ;
        for(int j = 1 ; j <= n ; j++)
        {
            if(!vis[j] && dis[j] > dis[pos] + ma[pos][j])
            {
                dis[j] = dis[pos] + ma[pos][j];
            }
        }
    }
}

int main()
{
    int to ;
    scanf("%d%d%d" , &n , &m , &to);
    int u , v , w ;
    memset(ma , INF , sizeof(ma));
    int ans = 0 ;
    for(int i = 1 ; i <= n ; i++)
        ma[i][i] = 0 ;

    for(int i = 1 ; i <= m ; i++)
    {
        scanf("%d%d%d" , &u , &v , &w);
        ma[u][v] = min(w , ma[u][v]);
    }
    
    Dijia(to);
    
    for(int i = 1 ; i <= n ; i++)
    {
        d[i] = dis[i];
    }
    
    for(int i = 1 ; i <= n ; i++)
    {
        for(int j = i + 1 ; j <= n ; j++)
        {
            int t ;
            t = ma[i][j];
            ma[i][j] = ma[j][i];
            ma[j][i] = t ;
        }
    }
    
    Dijia(to);
    
    for(int i = 1 ; i <= n ; i++)
    {
        ans = max(ans , d[i] + dis[i]);
    }

    printf("%d\n" , ans);


    return 0 ;
}

来源：https://www.cnblogs.com/nonames/p/11385747.html

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最短路径