Codeforces Round #349 (Div. 2) D. World Tour (最短路)

题目链接：http://codeforces.com/contest/667/problem/D

给你一个有向图，dis[i][j]表示i到j的最短路，让你求dis[u][i] + dis[i][j] + dis[j][v]的最大值，其中u i j v互不相同。

先用优先队列的dijkstra预处理出i到j的最短距离(n^2 logn)。(spfa也可以做)

然后枚举4个点的中间两个点i j，然后枚举与i相连节点的最短路dis[u][i](只要枚举最长的3个就行了)，接着枚举与j相连节点的最短路dis[j][v](同理也是枚举最长的3个就行了)，复杂度为9*n^2。

 #include <bits/stdc++.h>
 using namespace std;
 const int MAXN = ;
 typedef pair <int , int> P;
 struct data {
     int to , next;
 }edge[MAXN];
 int head[MAXN] , cnt , dis[MAXN][MAXN] , INF = 1e9;
 vector <P> G[MAXN] , RG[MAXN];

 inline void add(int u , int v) {
     edge[cnt].next = head[u];
     edge[cnt].to = v;
     head[u] = cnt++;
 }

 bool judge(int x1 , int x2 , int x3 , int x4) {
     if(x1 == x2 || x1 == x3 || x1 == x4 || x2 == x3 || x2 == x4 || x3 == x4)
         return false;
     return true;
 }

 void dijkstra(int s) {
     dis[s][s] = ;
     priority_queue <P , vector<P> , greater<P> > que;
     while(!que.empty()) {
         que.pop();
     }
     que.push(P( , s));
     while(!que.empty()) {
         P temp = que.top();
         que.pop();
         int u = temp.second;
         if(dis[s][u] < temp.first)
             continue;
         for(int i = head[u] ; ~i ; i = edge[i].next) {
             int v = edge[i].to;
             if(dis[s][v] > dis[s][u] + ) {
                 dis[s][v] = dis[s][u] + ;
                 que.push(P(dis[s][v] , v));
             }
         }
     }
 }

 int main()
 {
     memset(head , - , sizeof(head));
     cnt = ;
     int n , m , u , v;
     scanf("%d %d" , &n , &m);
     for(int i =  ; i <= n ; i++) {
         for(int j =  ; j <= n ; j++) {
             dis[i][j] = INF;
         }
     }
     while(m--) {
         scanf("%d %d" , &u , &v);
         add(u , v);
     }
     for(int i =  ; i <= n ; i++) {
         dijkstra(i);
         for(int j =  ; j <= n ; j++) {
             if(dis[i][j] != INF) {
                 G[i].push_back(P(dis[i][j] , j));
                 RG[j].push_back(P(dis[i][j] , i));
             }
         }
     }
     for(int i =  ; i <= n ; i++) {
         sort(G[i].begin() , G[i].end());
         sort(RG[i].begin() , RG[i].end());
     }
     int r1 , r2 , r3 , r4 , Max = -;
     for(int i =  ; i <= n ; i++) {
         for(int j =  ; j <= n ; j++) {
             if(i == j || dis[i][j] == INF)
                 continue;
             for(int x = RG[i].size() -  ; x >= max(int(RG[i].size() - ) , ) ; x--) {
                 for(int y = G[j].size() -  ; y >= max(int(G[j].size() - ) , ) ; y--) {
                     int xx = RG[i][x].second , yy = G[j][y].second;
                     if(dis[i][j] + dis[xx][i] + dis[j][yy] >= Max && judge(i , j , xx , yy)) {
                         r1 = xx , r2 = i , r3 = j , r4 = yy;
                         Max = dis[i][j] + dis[xx][i] + dis[j][yy];
                     }
                 }
             }
         }
     }
     printf("%d %d %d %d\n" , r1 , r2 , r3 , r4);
 }