正题

luogu题目链接:https://www.luogu.org/problemnew/show/CF666B

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题目大意

求4个点，使得这4个点按顺序最短路到达的长度最远。

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解题思路

用$bfs$求出每个点之间的最短路，然后对于每个点求出最远点，次远点，反最远点和反次远点。
之后枚举中间两个点，然后如果最远的相同就一个用次远，否则都用最远。

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N=5100,M=10100;
struct edge{int to,next;
}a[M];
int n,tot,m,f[N][N],maxs,ls[N],mark[4];
int ffar[N],ffar2[N],far[N],far2[N];
queue<int> q;
void adde(int x,int y)
{a[++tot].to=y;a[tot].next=ls[x];ls[x]=tot;
}
void bfs(int s)
{q.push(s);f[s][s]=0;while(!q.empty()){int x=q.front();q.pop();for(int i=ls[x];i;i=a[i].next){int y=a[i].to;if(f[s][y]<0){f[s][y]=f[s][x]+1;q.push(y);}}}
}
int main()
{memset(f,0xcf,sizeof(f));scanf("%d%d",&n,&m);for(int i=1;i<=m;i++){int x,y;scanf("%d%d",&x,&y);adde(x,y);}for(int i=1;i<=n;i++)bfs(i);for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(i==j) continue;if(f[i][j]>f[i][far[i]]){far2[i]=far[i];far[i]=j;}else if(f[i][j]>f[i][far2[i]])far2[i]=j;}}for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(i==j) continue;if(f[j][i]>f[ffar[i]][i]){ffar2[i]=ffar[i];ffar[i]=j;}else if(f[j][i]>f[ffar2[i]][i])ffar2[i]=j;}}for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(i!=j){if(!ffar[i]||!far[j]||ffar[i]==j||far[j]==i) continue;if(ffar[i]==far[j]){if(f[ffar2[i]][i]+f[j][far[j]]+f[i][j]>maxs&&ffar2[i]){maxs=f[ffar2[i]][i]+f[j][far[j]]+f[i][j];mark[0]=ffar2[i];mark[1]=i;mark[2]=j;mark[3]=far[j];}if(f[ffar[i]][i]+f[j][far2[j]]+f[i][j]>maxs&&far2[j]){maxs=f[ffar[i]][i]+f[j][far2[j]]+f[i][j];mark[0]=ffar[i];mark[1]=i;mark[2]=j;mark[3]=far2[j];}}elseif(f[ffar[i]][i]+f[j][far[j]]+f[i][j]>maxs){maxs=f[ffar[i]][i]+f[j][far[j]]+f[i][j];mark[0]=ffar[i];mark[1]=i;mark[2]=j;mark[3]=far[j];}}printf("%d %d %d %d",mark[0],mark[1],mark[2],mark[3]);
}