poj3268 Silver Cow Party（两次dijkstra）

https://vjudge.net/problem/POJ-3268

一开始floyd超时了。。

对正图定点求最短，对逆图定点求最短，得到任意点到定点的往返最短路。

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define INF 0x3f3f3f3f
typedef unsigned long long ll;
using namespace std;
int t[1010][1010], rt[1010][1010];
int d[1010], rd[1010], vis[1010];
int n, m, x, a, b, T;
void dijkstra(int dist[], int a[][1010])
{
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i <= n; i++){
        dist[i] = INF;
    }
    dist[x] = 0;
    for(int i = 1; i <= n; i++){
        int mini = INF, k = -1;
        for(int j = 1; j <= n; j++){
            if(!vis[j]&&mini > dist[j]){
                mini = dist[j];
                k = j;
            }
        }
        vis[k] = 1;
        for(int j = 1; j <= n; j++){
            if(!vis[j]&&dist[j] > dist[k]+a[k][j]){
                dist[j] = dist[k]+a[k][j];
            }
        }
    }
}
int main()
{
    cin >> n >> m >> x;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            t[i][j] = INF; rt[i][j] = INF;
        }
    }
    for(int i = 1; i <= m; i++){
        cin >> a >> b >> T;
        t[a][b] = T;//图 
        rt[b][a] = T;//逆图 
    }
    dijkstra(d, t);
    dijkstra(rd, rt);
    int maxm = -INF;
    for(int i = 1; i <= n; i++){
        if(d[i]!=INF&&rd[i]!=INF&&d[i]+rd[i] > maxm){
            maxm = d[i]+rd[i];
        }
    }
    cout << maxm << endl;
    return 0;
}