CodeForces-1281E Jeremy Bearimy

传送门：CodeForces-1281E

思路分析

对于每条边单独计算贡献，假设u–>v，u左侧有x个点，v右侧有y个点

1. 最大：尽量让这条边用很多次，那么最多能用min$(x,y)$次
2. 最小：让相邻的两两配对一定是最小的，如果y为奇数，那么x一定也是奇数，那么v右侧可以两两配对剩一个v，u左侧同理，所以u，v配对

样例输入

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

样例输出

15 33
6 6

AC代码

#include <functional>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <vector>
#include <string>
#include <cstdio>
#include <chrono>
#include <random>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) x.size()
#define lowbit(x) x&(-x)
#define all(x) x.begin(),x.end()
#define mem(a,b) memset(a,b,sizeof(a))
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);

using namespace std;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N=1e6+10;

vector<pair<ll,ll>>e[N];
ll mi,ma,n,siz[N];

void dfs(int u,int fa){
    siz[u]=1;
    for(auto x:e[u]){
        ll v=x.fi;
        ll w=x.se;
        if(v==fa) continue;
        dfs(v,u);
        siz[u]+=siz[v];
        if(siz[v]&1) mi+=w;
        ma+=min(siz[v],n-siz[v])*w;
    }
}
 

int main() {
    IOS;
#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int t;
    cin>>t;
    while(t--){
        cin>>n;
        n*=2;
        for(int i=1;i<=n;i++) e[i].clear();
        for(int i=1;i<n;i++){
            ll u,v,w;
            cin>>u>>v>>w;
            e[u].push_back(mp(v,w));
            e[v].push_back(mp(u,w));
        }
        mi=0,ma=0;
        dfs(1,1);
        cout<<mi<<" "<<ma<<endl;
    }




    return 0;
}