Codeforces Round # 614 (Div. 2) E (thinking, configuration, the DP)

Right side configuration, to start from 0 assignment side, the right side is the initial select a 0, the assignment of each node that contribution to both sides of a strand (chain inclusive) of the product number, the current time assignment Chain one endpoint continued an edge, the edge weight is assigned a value of +1 last time. DFS to find the number of chains on both sides of the junction point of the composition (comprising a chain end), then the enumeration endpoint DP.

 1 #define HAVE_STRUCT_TIMESPEC
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 vector<long long>edge[3007];
 5 long long cnt[3007][3007];
 6 long long fa[3007][3007];
 7 long long dp[3007][3007];
 8 void dfs(long long now,long longF, Long  Long the root) {
 . 9      CNT [the root] [now] = . 1 ; // start point node number and the link current point of the side opposite to the point (endpoint including chain) 
10      FA [the root] [now] = F; // the same start point and the current side-chain parent node F 
. 11      for (Auto IT: Edge [now]) {
 12 is          IF (IT == F)
 13 is              Continue ;
 14          DFS (IT, now, the root );
 15          CNT [the root] [now] + = CNT [the root] [IT]; // the results back to the upper node 
16      }
 . 17  }
 18 is  Long  Long Solve ( Long  Long L, Long  Longr) { // find the answers to the endpoint l and r as the maximum strand 
. 19      IF (r == l) // start and end points can not be the same as 
20 is          return  0 ;
 21 is      IF (DP [l] [r] == 0 ) // L and r are as first strand ends 
22 is          DP [L] [r] = CNT [L] [r] * CNT [r] [L] + max (Solve (L, FA [L] [ R & lt]), Solve (R & lt, FA [R & lt] [L])); // this is a result both of the chain section of the number of nodes (including dot chain segments) plus the product of the current dot chain remove an endpoint answer ( one of the new addition of two end points, the new value of the right side of the right side of the primary chain Max + 1'd) 
23 is      return DP [L] [R & lt];
 24  }
 25  int main () {
 26 is      iOS: : sync_with_stdio ( to false );
 27      cin.tie (NULL);
 28     cout.tie(NULL);
29     long long n;
30     cin>>n;
31     for(long long i=1;i<n;++i){
32         long long u,v;
33         cin>>u>>v;
34         edge[u].emplace_back(v);
35         edge[v].emplace_back(u);
36     }
37     for(long long i=1;i<=n;++i)//枚举链的起点
38         dfs(i,-1,i);//深搜
39      Long  Long ANS = 0 ;
 40      for ( Long  Long I = . 1 ; I <= n-; I ++) // starting enumeration chain 
41 is          for ( Long  Long J = . 1 ; J <= n-; J ++) // end enumeration chain 
42 is              ANS = max (ANS, Solve (I, J));
 43 is      COUT << ANS;
 44 is      return  0 ;
 45 }