Codeforces 667D World Tour (最短路＋枚举)

题意

另dis[u][v]表示u到v的最短路，求最长的dis[i][j]+dis[j][k]+dis[k][l]，输出i, j, k, l。

思路

先n*nlogn预处理出来所有点的最短路，然后给所有的dis[i][j]排个序保存起来，给所有的dis[k][l]排个序保存起来，然后就可以枚举所有的j和k，从itoj找最大的4条，ktol找最大的4条加起来就行了，找4条是因为为了防止这4个点出现重合比如j找到k,k找到i什么的，不过题目本身没什么坑点，基本上思路对就能过了。

代码

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define LL long long#define lowbit(x) ((x)&(-x))#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1|1#define MP(a, b) make_pair(a, b)const int INF = 0x3f3f3f3f;const int MOD = 1000000007;const int maxn = 1e5 + 10;const double eps = 1e-8;const double PI = acos(-1.0);typedef pair<int, int> pii;struct node{    int d, to;};bool cmp(node a, node b){    return a.d > b.d;}vector<int> G[3030];vector<node> itoj[3030], ktol[3030];int dis[3030][3030];int vis[3030];int n, m;void spfa(int src){    for (int i = 1; i <= n; i++)        dis[src][i] = INF;    memset(vis, 0, sizeof(vis));    queue<int> q;    q.push(src);    vis[src] = 1;    dis[src][src] = 0;    while (!q.empty())    {        int u = q.front();        q.pop();        for (int i = 0; i < G[u].size(); i++)        {            int v = G[u][i];            if (dis[src][v] > dis[src][u] + 1)            {                dis[src][v] = dis[src][u] + 1;                if (!vis[v])                {                    vis[v] = 1;                    q.push(v);                }            }        }    }}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    while (cin >> n >> m)    {        for (int i = 1; i <= n; i++)            G[i].clear();        for (int i = 0; i < m; i++)        {            int u, v;            cin >> u >> v;            G[u].push_back(v);        }        for (int i = 1; i <= n; i++)            spfa(i);        for (int i = 1; i <= n; i++)            itoj[i].clear(), ktol[i].clear();        for (int i = 1; i <= n; i++)            for (int j = 1; j <= n; j++)            {                if (i == j) continue;                if (dis[i][j] != INF)                {                    itoj[j].push_back((node){dis[i][j], i});                    ktol[i].push_back((node){dis[i][j], j});                }            }        for (int i = 1; i <= n; i++)        {            sort(itoj[i].begin(), itoj[i].end(), cmp);            sort(ktol[i].begin(), ktol[i].end(), cmp);        }        int ans1, ans2, ans3, ans4;        int mmax = -1;        for (int j = 1; j <= n; j++)            for (int k = 1; k <= n; k++)            {                if (j == k || dis[j][k] == INF) continue;                for (int i = 0; i < 4 && i < itoj[j].size(); i++)                {                    if (itoj[j][i].to == k) continue;                    for (int l = 0; l < 4 && l < ktol[k].size(); l++)                    {                        if (ktol[k][l].to == itoj[j][i].to || ktol[k][l].to == j) continue;                        int dd = itoj[j][i].d + ktol[k][l].d + dis[j][k];                        if (dd > mmax)                        {                            mmax = dd;                            ans1 = itoj[j][i].to;                            ans2 = j;                            ans3 = k;                            ans4 = ktol[k][l].to;                        }                    }                }            }        cout << ans1 << " " << ans2 << " " << ans3 << " " << ans4 << endl;    }    return 0;}