CF1060E Sergey and Subways 假的点分治

题目传送门：http://codeforces.com/problemset/problem/1060/D

题意：给出$N$个点的一棵树，现在将距离为$2$的点之间连一条边，求所有点对之间最短路的和，$N \leq 10^5$

---

一道树上乱搞题搞得我点分治调了一个小时$QAQ$

点分治中唯一需要注意的是：统计答案时因为长度为奇数的路线除二后要向上取整，所以要在计算时加上路径长度为奇数的路径数量

点分治都是模板题，会一个就全会了$qwq$

#include<bits/stdc++.h>
#define MAXN 200010
#define int long long
using namespace std;

struct Edge{
    int end , upEd;
}Ed[MAXN << 1];
long long head[MAXN] , size[MAXN] , N , minSize , minDir , nowSize , cntEd;
long long ans;
bool vis[MAXN];

inline void addEd(int a , int b){
    Ed[++cntEd].end = b;
     Ed[cntEd].upEd = head[a];
     head[a] = cntEd;
}

//求当前子树大小
void getNowSize(int dir){
     vis[dir] = 1;
     nowSize++;
     for(int i = head[dir] ; i ; i = Ed[i].upEd)
         if(!vis[Ed[i].end])
            getNowSize(Ed[i].end);
    vis[dir] = 0;
}

//求重心
void getZX(int dir){
    vis[dir] = size[dir] = 1;
    int maxSize = 0;
    for(int i = head[dir] ; i ; i = Ed[i].upEd)
        if(!vis[Ed[i].end]){
            getZX(Ed[i].end);
            size[dir] += size[Ed[i].end];
            maxSize = max(maxSize , size[Ed[i].end]);
        }
    maxSize = max(maxSize , nowSize - size[dir]);
    if(maxSize < minSize){
        minSize = maxSize;
        minDir = dir;
    }
    vis[dir] = 0;
}

//算答案
pair < long long , long long > calAns(int dir , int dep){
    vis[dir] = 1;
    ans += dep + 1 >> 1;
    nowSize++;
    pair < long long , long long > q = make_pair(dep , dep & 1);
    for(int i = head[dir] ; i ; i = Ed[i].upEd)
        if(!vis[Ed[i].end]){
            pair < long long , long long > t = calAns(Ed[i].end , dep + 1);
            q.first += t.first;
            q.second += t.second;
        }
    vis[dir] = 0;
    return q;
}

void work(int dir){
    nowSize = 0;
    getNowSize(dir);
    minSize = nowSize;
    getZX(dir);
    long long t = minDir , sum = 0 , culJi = 0 , culOu = 0;
    vis[t] = 1;
    nowSize = 0;
    for(int i = head[t] ; i ; i = Ed[i].upEd)
        if(!vis[Ed[i].end]){
            int k = nowSize;
            //注意答案统计！
            pair < long long , long long > t = calAns(Ed[i].end , 1);
            ans += (sum * (nowSize - k) + t.first * k + t.second * culOu + culJi * (nowSize - k - t.second)) >> 1;
            sum += t.first;
            culJi += t.second;
            culOu += nowSize - k - t.second;
        }
    for(int i = head[t] ; i ; i = Ed[i].upEd)
        if(!vis[Ed[i].end])
            work(Ed[i].end);
}

signed main(){
  ios::sync_with_stdio(0);
    cin >> N;
    for(int i = 1 ; i < N ; i++){
        int a , b;
        cin >> a >> b;
        addEd(a , b);
        addEd(b , a);
    }
    work(1);
    cout << ans;
    return 0;
}

 

转载于:https://www.cnblogs.com/Itst/p/9748459.html