World Tour CodeForces - 667D [暴力+bfs求解最短路]-白红宇

World Tour CodeForces - 667D [暴力+bfs求解最短路]

阅读量：532 次

发布时间：2019-03-08

本文共 5485 字，大约阅读时间需要 18 分钟。

A famous sculptor Cicasso goes to a world tour!

Well, it is not actually a world-wide. But not everyone should have the opportunity to see works of sculptor, shouldn't he? Otherwise there will be no any exclusivity. So Cicasso will entirely hold the world tour in his native country — Berland.

Cicasso is very devoted to his work and he wants to be distracted as little as possible. Therefore he will visit only four cities. These cities will be different, so no one could think that he has "favourites". Of course, to save money, he will chose the shortest paths between these cities. But as you have probably guessed, Cicasso is a weird person. Although he doesn't like to organize exhibitions, he likes to travel around the country and enjoy its scenery. So he wants the total distance which he will travel to be as large as possible. However, the sculptor is bad in planning, so he asks you for help.

There are n cities and m one-way roads in Berland. You have to choose four different cities, which Cicasso will visit and also determine the order in which he will visit them. So that the total distance he will travel, if he visits cities in your order, starting from the first city in your list, and ending in the last, choosing each time the shortest route between a pair of cities — will be the largest.

Note that intermediate routes may pass through the cities, which are assigned to the tour, as well as pass twice through the same city. For example, the tour can look like that: $\text{[math]}$ . Four cities in the order of visiting marked as overlines: [1, 5, 2, 4].

Note that Berland is a high-tech country. So using nanotechnologies all roads were altered so that they have the same length. For the same reason moving using regular cars is not very popular in the country, and it can happen that there are such pairs of cities, one of which generally can not be reached by car from the other one. However, Cicasso is very conservative and cannot travel without the car. Choose cities so that the sculptor can make the tour using only the automobile. It is guaranteed that it is always possible to do.

Input

In the first line there is a pair of integers n and m (4 ≤ n ≤ 3000, 3 ≤ m ≤ 5000) — a number of cities and one-way roads in Berland.

Each of the next m lines contains a pair of integers ui, vi (1 ≤ ui, vi ≤ n) — a one-way road from the city ui to the city vi. Note that ui and viare not required to be distinct. Moreover, it can be several one-way roads between the same pair of cities.

Output

Print four integers — numbers of cities which Cicasso will visit according to optimal choice of the route. Numbers of cities should be printed in the order that Cicasso will visit them. If there are multiple solutions, print any of them.

     Example 
       input 
        Copy 
8 91 22 33 44 14 55 66 77 88 5
       output 
2 1 8 7

Note

Let d(x, y) be the shortest distance between cities x and y. Then in the example d(2, 1) = 3, d(1, 8) = 7, d(8, 7) = 3. The total distance equals 13.

题意:相邻点的边权都是1,选择4个点,使得A->B->C->D的路径和最大并且保证A->B,B->C,C->D都是最短路

思路:n<=3000 . O(n^2)枚举第二个点i,第三个点j. 再用O(n^2)的bfs预处理出所有到其他点的最长长度. 至于为什么用BFS,因为权值都相等且为1. 变相的dijkstra,但是dijkstra的复杂度要到O(n^3). 因为四个点一定要不同,所以对于每个点要处理出前三远的点,包括反向和顺向

复杂度O(9*n^2)

#include 
   
    #define MOD 1000000007#define INF 0x3f3f3f3f#define bug cout << "bug" << endl#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long ll;typedef pair
    
      pii;const int MAX_N=3000+3;vector 
     
       edge[MAX_N];int n,m,dis[MAX_N][MAX_N];pii shun[MAX_N][5],ni[MAX_N][5];void updates(int u,int v){    shun[u][3]=shun[u][2];    shun[u][2]=shun[u][1];    shun[u][1]=make_pair(v,dis[u][v]);    return ;}void updaten(int u,int v){    int t=dis[u][v];    if(t>ni[v][1].second){        ni[v][3]=ni[v][2];        ni[v][2]=ni[v][1];        ni[v][1]=make_pair(u,dis[u][v]);    }    else if(t>ni[v][2].second){        ni[v][3]=ni[v][2];        ni[v][2]=make_pair(u,dis[u][v]);    }    else if(t>ni[v][3].second){        ni[v][3]=make_pair(u,dis[u][v]);    }    return ;}void bfs(int st){    int vis[MAX_N];    memset(vis,0,sizeof(vis));    dis[st][st]=0;    queue 
      
       q;    q.push(st);    while(!q.empty()){        int v=q.front();q.pop();        vis[v]=1;        for(int i=0;i
       
        > n >>m;    for(int i=1;i<=m;i++){        int u,v;        scanf("%d%d",&u,&v);        edge[u].push_back(v);    }    memset(dis,INF,sizeof(dis));    for(int i=1;i<=n;i++)   bfs(i);//    for(int i=1;i<=n;i++){//        printf("shun[%d][%d].second=%d\n",i,1,shun[i][1].second);//    printf("~%d\n",dis[1][i]);//    }    int node[50];    int mx=-INF;    for(int i=1;i<=n;i++){        for(int j=1;j<=n;j++){            if(i==j)    continue;            for(int k=1;k<=3;k++){                for(int l=1;l<=3;l++){                    if(!check(i,j,k,l)) continue;                    int ans=dis[i][j]+ni[i][k].second+shun[j][l].second;                    if(ans>INF) continue;                    if(ans>mx){                        mx=ans;                        node[1]=ni[i][k].first;                        node[2]=i;                        node[3]=j;                        node[4]=shun[j][l].first;//                         printf("%d %d %d %d ~%d\n",node[1],node[2],node[3],node[4],ans);                    }                }            }        }    }    printf("%d %d %d %d\n",node[1],node[2],node[3],node[4]);    return 0;}

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