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 = 5010;
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] = 0;
    priority_queue <P , vector<P> , greater<P> > que;
    while(!que.empty()) {
        que.pop();
    }
    que.push(P(0 , 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] + 1) {
                dis[s][v] = dis[s][u] + 1;
                que.push(P(dis[s][v] , v));
            }
        }
    }
}

int main()
{
    memset(head , -1 , sizeof(head));
    cnt = 0;
    int n , m , u , v;
    scanf("%d %d" , &n , &m);
    for(int i = 1 ; i <= n ; i++) {
        for(int j = 1 ; j <= n ; j++) {
            dis[i][j] = INF;
        }
    }
    while(m--) {
        scanf("%d %d" , &u , &v);
        add(u , v);
    }
    for(int i = 1 ; i <= n ; i++) {
        dijkstra(i);
        for(int j = 1 ; 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 = 1 ; i <= n ; i++) {
        sort(G[i].begin() , G[i].end());
        sort(RG[i].begin() , RG[i].end());
    }
    int r1 , r2 , r3 , r4 , Max = -1;
    for(int i = 1 ; i <= n ; i++) {
        for(int j = 1 ; j <= n ; j++) {
            if(i == j || dis[i][j] == INF)
                continue;
            for(int x = RG[i].size() - 1 ; x >= max(int(RG[i].size() - 4) , 0) ; x--) {
                for(int y = G[j].size() - 1 ; y >= max(int(G[j].size() - 4) , 0) ; 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
" , r1 , r2 , r3 , r4);
}