hdoj 6060(2017 Multi-University Training Contest
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题目链接:RXD and dividing
题目大意:有n-1个数2到n,现在需要你去切分这n-1个数为k个区间,区间可以为空,给出n-1条边,有边权,使得这些点组成一棵树,现在要你去算你分的每个区间的贡献,res
f代表要使得所有的点都能直接或者间接相连,问需要那些边的边权和(这个要求最小),总贡献尽量要求大,问最大贡献
题目思路:我们要想得到最大的答案,那么就要尽可能的去利用这些边,也就是尽可能重复计算这些边,那么我们可以想到的是,我们就要先去放叶子结点,这样访问的边才多,那么现在我们需要去做的事情就是把每条边的访问次数给算出来,而且其实一条边会算多少次跟某个点的所有子孙节点个数有关,就比如样例中,2号点有3个子孙节点, 那么2号点连接父节点的那条边会算3+1次。3号点有0个子孙节点,那么3号点连接父节点的那条边会算0+1次,然后就要去判断到底谁是父节点了,dfs的判断一下就好,算贡献的时候,还需要注意到不能超过k
#include <map>#include <set>#include <cmath>#include <queue>#include <stack>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 1e6+10;struct Node{ ll to,cost;};vector<Node>G[maxn];ll val[maxn],sz[maxn];void dfs(int u,int fa){ sz[u] = 1; for(int i = 0;i < G[u].size();i++){ int v = G[u][i].to; if(v == fa) continue; dfs(v,u); val[v] = G[u][i].cost; sz[u] += sz[v]; }}int main(){ ll m,k,u,v,cost; while(~scanf("%lld%lld",&m,&k)){ for(ll i = 0;i < maxn;i++) G[i].clear(); ll mm = m-1; while(mm--){ scanf("%lld%lld%lld",&u,&v,&cost); Node nd; nd.to = v,nd.cost = cost; G[u].push_back(nd); nd.to = u,nd.cost = cost; G[v].push_back(nd); } dfs(1,-1); ll sum = 0; for(ll i = 2;i <= m;i++) sum += min(sz[i],k)*val[i]; printf("%lld\n",sum); } return 0;}
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