【bzoj4530】[Bjoi2014]大融合 并查集+线段树合并

线段树合并好神啊，表示我这种傻逼只能想到树剖O(nlog^2n)做法
先把原树建出来，每次查询就等价于计算子节点的size*(父亲节点所在联通块的大小-子节点的size)
用并查集找到节点的祖先，维护子树size
这个东西可以用线段树合并来做，查询就是查询dfs序上的一段区间

好像LCT+启发式合并更快？

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<iostream>#include<algorithm>#define maxn 200010#define N 2000010using namespace std;struct yts{int op,x,y;}q[maxn];int n,T,in[maxn],out[maxn],tot,num,cnt;int root[maxn],lch[N],rch[N],size[N],g[maxn],f[maxn],dep[maxn];int head[maxn],to[maxn],next[maxn];char s[5];void addedge(int x,int y){num++;to[num]=y;next[num]=head[x];head[x]=num;}int find(int x){if (f[x]==x) return x;else return f[x]=find(f[x]);}void modify(int &i,int l,int r,int x){if (!i) i=++cnt;if (l==r) {size[i]=1;return;}int mid=(l+r)/2;if (x<=mid) modify(lch[i],l,mid,x);if (mid<x) modify(rch[i],mid+1,r,x);size[i]=size[lch[i]]+size[rch[i]];}void dfs(int x){in[x]=++tot;modify(root[x],1,n,in[x]);for (int p=head[x];p;p=next[p])if (to[p]!=g[x]){g[to[p]]=x;dep[to[p]]=dep[x]+1;dfs(to[p]);}out[x]=tot;}int merge(int x,int y){if (!x) return y;if (!y) return x;size[x]+=size[y];lch[x]=merge(lch[x],lch[y]);rch[x]=merge(rch[x],rch[y]);return x;}int query(int &root,int l,int r,int L,int R){if (!root) return 0;if (L<=l && r<=R) return size[root];int mid=(l+r)/2,ans=0;if (L<=mid) ans+=query(lch[root],l,mid,L,R);if (mid<R) ans+=query(rch[root],mid+1,r,L,R);return ans;}int main(){scanf("%d%d",&n,&T);for (int i=1;i<=T;i++){int x,y;scanf("%s%d%d",s,&x,&y);if (s[0]=='A') addedge(x,y),addedge(y,x);if (s[0]=='A') q[i].op=0,q[i].x=x,q[i].y=y;if (s[0]=='Q') q[i].op=1,q[i].x=x,q[i].y=y;}for (int i=1;i<=n;i++) if (!g[i]) dfs(i);for (int i=1;i<=n;i++) f[i]=i;for (int i=1;i<=T;i++){if (q[i].op==0){int x=find(q[i].x),y=find(q[i].y);if (dep[x]>dep[y]) swap(x,y);root[x]=merge(root[x],root[y]);f[y]=x;}else{int x=q[i].x,y=q[i].y;if (dep[x]<dep[y]) swap(x,y);int z=find(x);long long num=query(root[z],1,n,in[x],out[x]);printf("%lld\n",((long long)size[root[z]]-num)*num);}}return 0;}