Silver Cow Party dijkstra 邻接矩阵

题意:floyd 会tle ，只有靠杰哥了。题目的意思是让你从n个点到x点的每条最短路径中选取 某一点前往x点和返回原来的点所耗费的时间之和 最大的那一组，并输出最大值。
思路：迪杰斯特拉搞一搞，这里会用到转置矩阵，从而实现求出前往x点和返回x点两条路径。（稠密图用邻接矩阵！！）

/**
* From:
* Qingdao Agricultural University
* Created by XiangwangAcmer
* Date : 2019-10-04-22.17.13
* Talk is cheap.Show me your code.
*/
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<cmath>
#include<cctype>
#include<stack>
#include<map>
#include<string>
#include<cstdlib>
#define ll long long
using namespace std;
const ll maxn = 1010;
const ll minn = 1e9 + 5;
const ll mod = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LIMIT = 4294967295LL;
int d[maxn][maxn], n;
int dis[maxn], dis1[maxn];
void dijkstra(int x) {
    int vis[maxn], min, next = x;
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i <= n; ++i)
        dis[i] = d[x][i];
    vis[x] = 1;
    for(int i = 2; i <= n; ++i) {
        min = INF;
        for(int j = 1; j <= n; ++j) {
            if(!vis[j] && dis[j] < min) {
                min = dis[j];
                next = j;
            }
        }
        vis[next] = 1;
        for(int j = 1; j <= n; ++j) {
            if(!vis[j] && dis[j] > dis[next] + d[next][j])
                dis[j] = dis[next] + d[next][j];
        }
    }
}

int main() {
    int m, w,x;
    int maxnn=-minn;
    cin >> n >> m >> x;
    memset(d, INF, sizeof d);
    for(int i = 1; i <= n; ++i)
        d[i][i] = 0;
    int a, b;
    while(m--) {
        cin >> a >> b >> w;
        d[a][b] = w;
    }
    dijkstra(x);
    int ans = 0;
    for(int i = 1; i <= n; i++)
        if(i != x)
            dis1[i] = dis[i];
    for(int i = 1; i <= n; i++)
        for(int j = i + 1; j <= n; j++) {
           int t = d[i][j];
            d[i][j] = d[j][i];
            d[j][i] = t;
        }
    dijkstra(x);
    for(int i=1;i<=n;i++)
        if(i!=x)
        maxnn=max(dis1[i]+dis[i],maxnn);
    cout << maxnn<< endl;
    return 0;
}