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

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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 ;}

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