CF-915-F-并查集

合集 - Codeforces(31)

1.CF-55-D-数位DP2024-01-232.CF-87-D-并查集2024-01-133.CF-91-B-单调栈+二分2024-01-234.CF-240-F-线段树2024-01-215.CF-242-E-线段树2024-01-206.CF-292-D-并查集2024-01-237.CF-282-E-Trie2024-01-178.CF-342-E-分块2024-01-129.CF-375-D-启发式合并2024-01-1210.CF-431-D-二分+数位DP2024-01-2411.CF-461-B-树形DP2024-01-2212.CF-514-D-单调队列2024-01-1413.CF-522-D-线段树2024-05-0514.CF-570-D-启发式合并2024-01-2115.CF-600-E-启发式合并2024-05-0416.CF-613-D-虚树2024-01-1317.CF-620-E-DFS序+线段树2024-01-1718.CF-675-E-RMQ优化DP2024-01-1619.CF-702-E-倍增2024-01-1720.CF-797-E-根号分治2024-05-0421.CF-817-E-Trie2024-01-1622.CF-877-E-dfs序+线段树2024-05-05

23.CF-915-F-并查集2024-01-21

24.CF-938-D-最短路2024-05-1025.CF-1051-F-最短路+最小生成树2024-01-1726.CF-1184-E3-最小生成树+倍增+并查集2024-01-2427.CF-1304-E-倍增LCA+思维2024-01-2528.CF-1399-E2-优先队列2024-01-2229.CF-1515-F-思维2024-02-2430.CF-1831-E-卡特兰数+异或哈希+差分2024-01-2231.CF-1921-F-根号分治2024-01-24

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915-F 题目大意

给定一棵$n$个节点的树，节点带权，设函数$I(x,y)$等于点$x$到点$y$的路径上最大的点权与最小的点权之差。求：

$$\sum _{i=1}^n\sum _{j=i}^{n}I(i,j)$$

---

Solution

令函数$F(i,j)$等于点$x$到点$y$的路径上最大的点权，函数$G(i,j)$等于点$x$到点$y$的路径上最小的点权。所求的式子可转换为：

$$\sum _{i=1}^n\sum _{j=i}^{n}F(i,j)-\sum _{i=1}^n\sum _{j=i}^{n}G(i,j)$$

先求所有点对的$F(i,j)$之和，按点权从小到大枚举点$x$，枚举$x$连接的连通块，这些连通块的大小两两相乘，从而能够得到点$x$作为最大值的贡献，枚举的同时用并查集合并这些边。$G(i,j)$之和的求法相同。

时间复杂度$O(nlogn)$，瓶颈在排序上。

#include<bits/stdc++.h>
using namespace std;
using ll=long long;

template<typename T>
struct DSU{
	int n;
	vector<T> p,siz;
	DSU(int _n):p(_n+1),siz(_n+1),n(_n){
		iota(p.begin(),p.end(),0);
		for(int i=0;i<=n;i++) siz[i]=1;
	}
	T findd(T x){
		return p[x]==x?x:p[x]=findd(p[x]);
	}
	void unionn(T x,T y){
		x=findd(x),y=findd(y);
		if(x==y) return;
		if(siz[x]>siz[y]) swap(x,y);
		p[x]=y;
		siz[y]+=siz[x];
	}
};

void solve(){
    int n;
    cin>>n;
    vector<int> a(n);
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    vector<vector<int>> e(n);
    for(int i=1;i<n;i++){
        int x,y;
        cin>>x>>y;
        x--,y--;
        e[x].push_back(y);
        e[y].push_back(x);
    }
    vector<int> ids(n),vis(n);
    iota(ids.begin(),ids.end(),0);
    sort(ids.begin(),ids.end(),[&](const auto &i,const auto &j){
        return a[i]<a[j];
    });
    ll ans=0;
    DSU<int> mx(n),mn(n);
    for(int i=0;i<n;i++){
        int x=ids[i];
        int cnt=1;
        for(auto y:e[x]){
            if(!vis[y]) continue;
            ans+=1LL*cnt*mx.siz[mx.findd(y)]*a[x];
            cnt+=mx.siz[mx.findd(y)];
            mx.unionn(x,y);
        }
        vis[x]=1;
    }
    vis.assign(n,0);
    for(int i=n-1;~i;i--){
        int x=ids[i];
        int cnt=1;
        for(auto y:e[x]){
            if(!vis[y]) continue;
            ans-=1LL*cnt*mn.siz[mn.findd(y)]*a[x];
            cnt+=mn.siz[mn.findd(y)];
            mn.unionn(x,y);
        }
        vis[x]=1;
    }
    cout<<ans;
}

int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}