（其实原题强制在线，要用主席树）
代码：
#include<bits/stdc++.h> #define reg register int #define il inline #define numb (ch^'0') using namespace std; typedef long long ll; il void rd(int &x){ char ch;x=0;bool fl=false; while(!isdigit(ch=getchar()))(ch=='-')&&(fl=true); for(x=numb;isdigit(ch=getchar());x=x*10+numb); (fl==true)&&(x=-x); } namespace Miracle{ const int M=400000+5; const int N=400000+5; const int inf=0x3f3f3f3f; int n,m,Q; struct edge{ int x,y; int val; }b[M]; bool cmp0(edge a,edge b){ return a.val<b.val; } bool cmp1(edge a,edge b){ return a.val>b.val; } int fafa[2*N]; struct kruskal{ struct node{ int nxt,to; }e[2*N]; int hd[2*N],cnt; int fin(int x){ return fafa[x]==x?x:fafa[x]=fin(fafa[x]); } void add(int x,int y){ e[++cnt].nxt=hd[x]; e[cnt].to=y; hd[x]=cnt; } int tot; void build(int typ){ for(reg i=1;i<=n;++i){ if(typ) val[i]=inf; else val[i]=-inf; } tot=n; for(reg i=1;i<=m;++i){ int k1=fin(b[i].x),k2=fin(b[i].y); // cout<<" edge "<<b[i].x<<" "<<b[i].y<<" :: "<<k1<<" "<<k2<<endl; if(k1!=k2){ ++tot; fafa[tot]=tot; fafa[k1]=tot; fafa[k2]=tot; val[tot]=b[i].val; add(tot,k1); add(tot,k2); } } } int l[N],r[N]; int val[N]; int fa[N][20]; int df; void dfs(int x){ // cout<<" xx "<<x<<endl; int son=0; r[x]=-inf;l[x]=inf; for(reg i=hd[x];i;i=e[i].nxt){ int y=e[i].to; ++son; dfs(y); fa[y][0]=x; r[x]=max(r[x],r[y]); l[x]=min(l[x],l[y]); } if(!son){ l[x]=r[x]=++df; } } void pre(){ dfs(tot); for(reg j=1;j<=19;++j){ for(reg i=1;i<=tot;++i){ fa[i][j]=fa[fa[i][j-1]][j-1]; } } } int fin(int x,int lim,int typ){//beizeng go val int p=x; if(!typ){//go <=lim for(reg j=19;j>=0;--j){ if(fa[p][j]){ if(val[fa[p][j]]<=lim) p=fa[p][j]; } } return p; }else{//go >=lim for(reg j=19;j>=0;--j){ if(fa[p][j]){ if(val[fa[p][j]]>=lim) p=fa[p][j]; } } return p; } } }kt[2];//0:min tree;1:max tree; int num; struct po{ int x,y; bool friend operator <(po a,po b){ return a.x<b.x; } }p[N]; int ans[N]; int tot; struct que{ int id,x,typ,y1,y2; bool friend operator <(que a,que b){ return a.x<b.x; } }q[N*2]; struct binarytree{ int f[N]; void upda(int x){ for(;x<=n;x+=x&(-x)) f[x]++; } int query(int x){ int ret=0; for(;x;x-=x&(-x)) ret+=f[x]; return ret; } }t; int main(){ rd(n);rd(m);rd(Q); for(reg i=1;i<=m;++i){ rd(b[i].x);rd(b[i].y); ++b[i].x;++b[i].y;//warning!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! b[i].val=max(b[i].x,b[i].y); } sort(b+1,b+m+1,cmp0); for(reg i=1;i<=2*n;++i){ fafa[i]=i; } kt[0].build(0); kt[0].pre(); // cout<<" after build small "<<endl; for(reg i=1;i<=m;++i){ b[i].val=min(b[i].x,b[i].y); } sort(b+1,b+m+1,cmp1); for(reg i=1;i<=2*n;++i){ fafa[i]=i; } kt[1].build(1); kt[1].pre(); int st,nd,L,R; for(reg i=1;i<=Q;++i){ rd(st);rd(nd);rd(L);rd(R); ++L;++R; ++st;++nd;//warning!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! int ptr=kt[1].fin(st,L,1); q[++tot].id=i; q[tot].y1=kt[1].l[ptr]; q[tot].y2=kt[1].r[ptr]; q[++tot].id=i; q[tot].y1=kt[1].l[ptr]; q[tot].y2=kt[1].r[ptr]; ptr=kt[0].fin(nd,R,0); q[tot-1].x=kt[0].l[ptr]-1; q[tot].x=kt[0].r[ptr]; q[tot-1].typ=-1; q[tot].typ=1; } sort(q+1,q+tot+1); for(reg i=1;i<=n;++i){ p[i].x=kt[0].l[i]; p[i].y=kt[1].l[i]; } sort(p+1,p+n+1); int ptp=1,ptq=1; for(reg i=1;i<=n;++i){ while(ptp<=n&&p[ptp].x<i) ++ptp; if(p[ptp].x==i){ while(ptp<=n&&p[ptp].x==i){ t.upda(p[ptp].y); ++ptp; } } while(ptq<=tot&&q[ptq].x<i) ++ptq; if(q[ptq].x==i){ while(ptq<=tot&&q[ptq].x==i){ ans[q[ptq].id]+=q[ptq].typ*(t.query(q[ptq].y2)-t.query(q[ptq].y1-1)); ++ptq; } } } for(reg i=1;i<=Q;++i){ puts(ans[i]?"1":"0"); } return 0; } } signed main(){ Miracle::main(); return 0; } /* Author: *Miracle* Date: 2019/2/10 15:50:00 */
总结：
kruskal重构树，就是考虑在经过边权在一定范围内到达的区域的点的情况
这里就是简单查询连通性
两个重构树判交的“二维数点”问题的转化很巧妙！