LuoGuP1351联合权值

LuoGuP1351联合权值
并不是很难的题目.
你考虑枚举中间点,然后所有与它相连的点都满足形成权值的条件.
所以我们就直接枚举中间点,取权值最大的两个相乘更新答案.
然后统计所有和它相连的点的权值和,每次统计总权值的时候加上(相连的点的权值和-它自身的权值)*它自身的权值即可.

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
#define int long long
#define pb push_back

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

const int mod = 1e4 + 7 ;
const int N = 2e5 + 100 ;
vector < int > G[N] ;
int n , v[N] , sum[N] ;
int maxn , ans ;

inline bool cmp (int a , int b) { return v[a] > v[b] ; }

signed main (int argc , char * argv[]) {
    n = rint () ; rep ( i , 2 , n ) {
        int u = rint () , v = rint () ;
        G[u].pb ( v ) ; G[v].pb ( u ) ;
    }
    rep ( i , 1 , n ) v[i] = rint () ;
    rep ( i , 1 , n ) sort ( G[i].begin () , G[i].end () , cmp ) ;
    rep ( i , 1 , n ) {
        if ( (int)G[i].size () < 2 ) continue ; int sum = 0 ;
        for (int j = 0 ; j < (int)G[i].size () ; ++ j) sum += v[G[i][j]] ;
        maxn = max ( maxn , v[G[i][0]] * v[G[i][1]] ) ;
        for (int j = 0 ; j < (int)G[i].size () ; ++ j) {
            int k = G[i][j] , m = (int)G[i].size () ;
            ans = ( ans + v[k] * ( sum - v[k] ) % mod ) % mod ;
        }
    }
    printf ("%lld %lld
" , maxn , ans ) ;
    system ("pause") ; return 0 ;
}