洛谷 P4219 [BJOI2014]大融合

查询，就相当于先删去这条边，然后查询边的两个端点所在连通块大小，乘起来得到答案，然后再把边加回去

可以用线段树分治做

 #pragma GCC optimize("Ofast")
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<vector>
 #include<map>
 using namespace std;
 #define fi first
 #define se second
 #define mp make_pair
 #define pb push_back
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int,int> pi;
 #define N 100100
 struct Q
 {
     int type,x,y,l,r,num;
 }q[N*];
 ll ans[N];
 int nq,n,qq,tm,an;
 char ss[];
 int fa[N],sz[N],dp[N];
 map<pi,int> ma;
 int find(int x)
 {
     for(;x!=fa[x];x=fa[x]);
     return x;
 }
 int bx[N*],bfa[N*],bsz[N*],bdp[N*],mem;
 void unionn(int x,int y)
 {
     x=find(x),y=find(y);
     ++mem;bx[mem]=x;bfa[mem]=fa[x];bsz[mem]=sz[x];bdp[mem]=dp[x];
     ++mem;bx[mem]=y;bfa[mem]=fa[y];bsz[mem]=sz[y];bdp[mem]=dp[y];
     if(dp[x]<dp[y])    {fa[x]=y;sz[y]+=sz[x];}
     else
     {
         fa[y]=x;sz[x]+=sz[y];
         if(dp[x]==dp[y])    dp[x]++;
     }
 }
 void backn(int nn)
 {
     for(int i=;i<=nn;i++)
     {
         fa[bx[mem]]=bfa[mem];sz[bx[mem]]=bsz[mem];dp[bx[mem]]=bdp[mem];--mem;
         fa[bx[mem]]=bfa[mem];sz[bx[mem]]=bsz[mem];dp[bx[mem]]=bdp[mem];--mem;
     }
 }
 void solve(int pl,int pr,int l,int r)//[l,r]为线段树上区间
 {
     if(pl>pr)    return;
     int i,nn=;
     if(l==r)
     {
         for(i=pl;i<=pr;i++)
             if(q[i].type==)
                 unionn(q[i].x,q[i].y),++nn;
         for(i=pl;i<=pr;i++)
             if(q[i].type==)
                 ans[q[i].num]=ll(sz[find(q[i].x)])*sz[find(q[i].y)];
         backn(nn);
         return;
     }
     int mid=l+((r-l)>>),p=pl-;
     for(i=pl;i<=pr;i++)
     {
         if(q[i].l<=l&&r<=q[i].r)    unionn(q[i].x,q[i].y),++nn;
         else    if(q[i].l<=mid)    swap(q[++p],q[i]);
     }
     solve(pl,p,l,mid);p=pl-;
     for(i=pl;i<=pr;i++)
     {
         if((q[i].l>l||r>q[i].r)&&mid<q[i].r)    swap(q[++p],q[i]);
     }
     solve(pl,p,mid+,r);
     backn(nn);
 }
 int main()
 {
     int i,x,y;
     scanf("%d%d",&n,&qq);
     for(i=;i<=qq;i++)
     {
         scanf("%s%d%d",ss,&x,&y);
         if(x>y)    swap(x,y);
         if(ss[]=='A')
         {
             ma[mp(x,y)]=++tm;
         }
         else
         {
             q[++nq]=(Q){,x,y,ma[mp(x,y)],++tm,};
             ++tm;q[++nq]=(Q){,x,y,tm,tm,++an};
             ma[mp(x,y)]=++tm;
         }
     }
     for(auto xx:ma)    q[++nq]=(Q){,xx.fi.fi,xx.fi.se,xx.se,tm,};
     for(i=;i<=n;i++)    fa[i]=i,sz[i]=;
     solve(,nq,,tm);
     for(i=;i<=an;i++)    printf("%lld\n",ans[i]);
     return ;
 }