牛客网暑期ACM多校训练营（第六场）G

https://www.nowcoder.com/acm/contest/144/G

链接：https://www.nowcoder.com/acm/contest/144/G
来源：牛客网

In Viridian forest there is a tree T formed by N nodes, each edge on which has a positive weight.

There is an undirected graph G generated from tree T, which contains N nodes and $\displaystyle\binom{N}{2}$ undirected edges, where the capacity of the edge between u and v equals to the distance between them on the tree T.

Given the tree T, Pikachu wants to calculate the sum of the max flow between every two nodes in G, there are $\displaystyle\binom{N}{2}$ different pairs of nodes should be counted. Could you help him?

输入描述:

The input starts with one line containing exactly one integer t, which is the number of test cases.

For each test case, the first line contains one integer N, indicating the size of the tree T.

Then followed by N - 1 lines, each consists of three integers u

, v

 and w

, representing the two nodes connected by the i-th edge and the weight of the i-th edge.


- 1 ≤ t ≤ 10.
- 2 ≤ N ≤ 10

⁵

.

- 1 ≤ w_i≤ 1000.

$1 \le u_i \not= v_i \le N.$

输出描述:

For each test case, output one line containing "Case #x: y", where x is the test case number (starting from 1) and y is the sum of the maximum flow between every two nodes in G.

示例1

输入

输出

Case #1: 7
Case #2: 72


题解

我感觉他说的显然那句话显然是错的啊但是答案貌似是他说的那样啊找不到反例啊不知道对不对啊没人解答啊

就当做求树上所有点的距离之和来做吧

代码

#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("\n");
#define debug(a,b) cout<<a<<" "<<b<<" "<<endl;
using namespace std;
typedef long long ll;
const ll maxn=1e5+50,inf=1e18;
const ll mod=1e15;
ll f[maxn],h[maxn],sz[maxn];
vector<pair<ll,ll>> g[maxn];
int n;
void dfs1(int x,int y)
{
    sz[x]=1;
    f[x]=0;
    for(int i=0;i<g[x].size();i++)
    {
        int u=g[x][i].fi;
        ll w=g[x][i].se;
        if(u==y)continue;
        dfs1(u,x);
        sz[x]+=sz[u];
        f[x]+=f[u]+sz[u]*w;
    }
}
void dfs2(int x,int y)
{
    for(int i=0;i<g[x].size();i++)
    {
        int u=g[x][i].fi;
        ll w=g[x][i].se;
        if(u==y)continue;
        h[u]=h[x]+f[x]-f[u]-sz[u]*w+(n-sz[u])*w;
        dfs2(u,x);
    }
    f[x]+=h[x];
}
int main()
{
    int t,kase=1;
    cin>>t;
    while(t--)
    {
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            h[i]=f[i]=sz[i]=0;
            g[i].clear();
        }
        for(int i=1;i<n;i++)
        {
            int u,v,w;
            cin>>u>>v>>w;
            g[u].pb(mp(v,w));
            g[v].pb(mp(u,w));
        }
        dfs1(1,0);
        dfs2(1,0);
        ll ans1=0,ans2=0;
        sort(f+1,f+1+n);
        for(int i=1;i<=n;i++)
        {
            ll temp=f[i]*(n-i); //第一小的贡献了n-1次  以此类推
            ans2+=temp%mod;
            ans1+=temp/mod;
            if(ans2>=mod)      //相加过程会爆ll 所以这样处理一下 分成前后15位
            {
                ans2-=mod;
                ans1++;
            }
        }
        printf("Case #%d: ",kase++);
        if(ans1)
            printf("%lld%015lld\n",ans1,ans2);
        else
            printf("%lld\n",ans2);
    }
}