51NOD1681 公共祖先 【dfs序+树状数组】

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首先求一遍树1的dfs序,线段树维护数组c[i]
dfs遍历树2
当进入u点时
①: subNum=∑out[u]i=in[u]c[i]
②: c[in[u]]+=1
③: 遍历u的所有孩子
④: 以u为树根的2个子树包含的公共点数=∑out[u]i=in[u]c[i]−subNum

#include<stdio.h>#include<bits/stdc++.h>#define ll long long#define lowbit(x) ((x)&(-x))using namespace std;const int N = 1e5 +5 ;int in[N],out[N];int inDeg1[N],inDeg2[N];struct BIT{    ll c[N];    int n;    void init(int n){        this->n=n;        fill(c,c+n+1,0);    }    void add(int x,ll val){        while(x<=n){            c[x]+=val;            x+=lowbit(x);        }    }    ll sum(int x){        ll ans=0;        while(x){            ans+=c[x];            x-=lowbit(x);        }        return ans;    }}bit;vector<int>G1[N],G2[N];void init(int n){    for(int i=0;i<=n;++i){        G1[i].clear();        G2[i].clear();        inDeg1[i]=inDeg2[i]=0;    }    bit.init(n);}void dfs(int u,int&t){    in[u]=++t;    for(int v:G1[u]){        dfs(v,t);    }    out[u]=t;}ll f(int n){    return (ll)n*(n-1)/2;}ll ans=0;void dfs2(int u){    ll subNum=bit.sum(out[u])-bit.sum(in[u]-1);    bit.add(in[u],1);    for(int v:G2[u]){        dfs2(v);    }    ll num=bit.sum(out[u])-bit.sum(in[u]-1)-subNum;    ans+=f(num-1);}ll slove(int n){    ans=0;    for(int i=1,t=0;i<=n;++i){        if(inDeg1[i]==0){            dfs(i,t);        }    }    for(int i=1;i<=n;++i){        if(inDeg2[i]==0){            dfs2(i);        }    }    return ans;}int main(){    int n;    while(~scanf("%d",&n)){        init(n);        for(int i=0;i<n-1;++i){            int u,v;            scanf("%d%d",&u,&v);            G1[u].push_back(v);            ++inDeg1[v];        }        for(int i=0;i<n-1;++i){            int u,v;            scanf("%d%d",&u,&v);            G2[u].push_back(v);            ++inDeg2[v];        }        printf("%lld\n",slove(n));    }    return 0;}