HDU 6060 RXD and dividing（dfs 思维）

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1893    Accepted Submission(s): 809

Problem Description

RXD has a tree ) is equal to the minimal cost

Input

There are several test cases, please keep reading until EOF.
For each test case, the first line consists of 2 integer 100.

Output

For each test case, output an integer, which means the answer.

Sample Input

5 4 1 2 3 2 3 4 2 4 5 2 5 6

Sample Output

27

Source

2017 Multi-University Training Contest - Team 3

【题意】给你一棵树，将节点[2,n]最多分为k份，再将1节点加入到每一份，将每一份的节点连接起来，权值之和加入ans，求最大化ans。

【分析】咋一看，发现很难，不会...但看了题解后仔细一想，就是个傻逼题啊...每一个 节点与他父亲节点之间的权值的贡献就是他子树分成的份数，那我们就最大化这个份   数就是了...

  

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 1e6+50;;
const int M = 255;
const int mod = 19260817;
const int mo=123;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
int n,k,cas;
ll ans;
int sz[N];
vector<pii>edg[N];
void dfs(int u,int fa){
    sz[u]=1;
    for(auto e : edg[u]){
        int v=e.first;
        int w=e.second;
        if(v==fa)continue;
        dfs(v,u);
        sz[u]+=sz[v];
        ans+=1LL*w*min(sz[v],k);
    }
}
int main(){
    //int T;
    //scanf("%d",&T);
    while(~scanf("%d%d",&n,&k)){
        for(int i=1;i<=n;i++)edg[i].clear();
        for(int i=1,u,v,w;i<n;i++){
            scanf("%d%d%d",&u,&v,&w);
            edg[u].pb(mp(v,w));edg[v].pb(mp(u,w));
        }
        ans=0;
        dfs(1,0);
        printf("%lld\n",ans);
    }
    return 0;
}