【codeforces】gym 101137 K - Knights of the Old Republic【用最小生成树对图做集合dp】

题目链接：【codeforces】gym 101137 K - Knights of the Old Republic

考虑对图集合dp，一个连通块的dp值为两个连通块的值的和或者强制加一条新边后的最小值，取个最小值（边从小到大枚举，则强制加一条最大的边会导致连通块内较小的边一定都选，则会构成一个生成树）。用kruskal实现这个dp过程即可。

#include <bits/stdc++.h>
using namespace std ;

typedef long long LL ;

const int MAXN = 300005 ;

struct Edge {
    int u , v , c ;
    bool operator < ( const Edge& a ) const {
        return c < a.c ;
    }
} ;

Edge E[MAXN] ;
int a[MAXN] , b[MAXN] , p[MAXN] ;
LL sum[MAXN] ;
int n , m ;

int F ( int x ) {
    return p[x] == x ? x : ( p[x] = F ( p[x] ) ) ;
}

void solve () {
    for ( int i = 1 ; i <= n ; ++ i ) {
        p[i] = i ;
    }
    for ( int i = 1 ; i <= n ; ++ i ) {
        scanf ( "%d%d" , &a[i] , &b[i] ) ;
        sum[i] = 1LL * a[i] * b[i] ;
    }
    for ( int i = 0 ; i < m ; ++ i ) {
        scanf ( "%d%d%d" , &E[i].u , &E[i].v , &E[i].c ) ;
    }
    sort ( E , E + m ) ;
    for ( int i = 0 ; i < m ; ++ i ) {
        int x = F ( E[i].u ) ;
        int y = F ( E[i].v ) ;
        if ( x == y ) continue ;
        int na = max ( max ( a[x] , a[y] ) , E[i].c ) ;
        int nb = min ( b[x] , b[y] ) ;
        LL tmp = 1LL * na * nb ;
        sum[y] = min ( sum[x] + sum[y] , tmp ) ;
        p[x] = y ;
        a[y] = na ;
        b[y] = nb ;
    }
    LL ans = 0 ;
    for ( int i = 1 ; i <= n ; ++ i ) {
        if ( F ( i ) == i ) ans += sum[i] ;
    }
    printf ( "%lld\n" , ans ) ;
}

int main () {
    while ( ~scanf ( "%d%d" , &n , &m ) ) solve () ;
    return 0 ;
}