[整体二分 并查集]Atcoder AGC002 D. Stamp Rally

每个询问就是求最小的 x，使得保留编号小于等于 x 的边后，和 ai,bi 相连的点大于等于 zi

这个东西整体二分一下，用按秩合并的并查集维护一下就好了吧

#include <cstdio>#include <iostream>#include <algorithm>#define fi first#define se secondusing namespace std;typedef pair<int,int> par;const int N=100010;int n,m,q,fa[N],ans[N],size[N],rk[N];par e[N];struct Qry{  int x,y,z,g;}Q[N],a[N],b[N];inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(); x=0;  for(;c>'9'||c<'0';c=nc());for(;c>='0'&&c<='9';x=x*10+c-'0',c=nc());}int Gfat(int x){  return fa[x]==x?x:Gfat(fa[x]);}struct sta{  int x,rk;}S[N];int t;inline void Union(int x,int y){  x=Gfat(x); y=Gfat(y);  if(x==y) return ;  if(rk[x]>rk[y]) swap(x,y);  fa[x]=y; size[y]+=size[x]; S[++t]=(sta){x,rk[y]};  if(rk[x]==rk[y]) rk[y]++;}inline void Back(){  int x=S[t].x; size[fa[x]]-=size[x]; rk[fa[x]]=S[t].rk;  fa[x]=x; t--;}void solve(int l,int r,int L,int R){  if(l>r) return ;  int mid=l+r>>1,t1=0,t2=0,cur=t;  for(int i=l;i<=mid;i++)    Union(e[i].fi,e[i].se);  for(int i=L;i<=R;i++){    int x=Gfat(Q[i].x),y=Gfat(Q[i].y);    if(x==y){      if(size[x]>=Q[i].z)    a[++t1]=Q[i],ans[Q[i].g]=mid;      else    b[++t2]=Q[i];    }    else{      if(size[x]+size[y]>=Q[i].z)    a[++t1]=Q[i],ans[Q[i].g]=mid;      else    b[++t2]=Q[i];    }  }  for(int i=1;i<=t1;i++) Q[L+i-1]=a[i];  for(int i=1;i<=t2;i++) Q[L+t1+i-1]=b[i];  solve(mid+1,r,L+t1,R);  while(t>cur) Back();  solve(l,mid-1,L,L+t1-1);}int main(){  freopen("1.in","r",stdin);  freopen("1.out","w",stdout);  read(n); read(m);  for(int i=1;i<=n;i++)    fa[i]=i,size[i]=rk[i]=1;  for(int i=1;i<=m;i++)    read(e[i].fi),read(e[i].se);  read(q);  for(int i=1;i<=q;i++)    read(Q[i].x),read(Q[i].y),read(Q[i].z),Q[i].g=i;  solve(1,m,1,q);  for(int i=1;i<=q;i++) printf("%d\n",ans[i]);  return 0;}