数据结构（树套树）：ZJOI 2013 K大数查询

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <ctime>
 5 using namespace std;
 6 const int maxn=50010;
 7 const int maxm=20000010;
 8 int rt[maxn*4],tot,n,m;
 9 int ch[maxm][2],sum[maxm],tag[maxm];
10 
11 void Push_up(int x){
12     sum[x]=sum[ch[x][0]]+sum[ch[x][1]];
13 }
14 
15 void Add(int &x,int l,int r,int d){
16     if(!x)x=++tot;
17     sum[x]+=(r-l+1)*d;
18     tag[x]+=d;
19 }
20 
21 void Push_down(int x,int l,int r){
22     if(tag[x]&&l!=r){
23         int mid=(l+r)>>1;
24         if(!ch[x][0])ch[x][0]=++tot;
25         if(!ch[x][1])ch[x][1]=++tot;
26         sum[ch[x][0]]+=(mid-l+1)*tag[x];tag[ch[x][0]]+=tag[x];
27         sum[ch[x][1]]+=(r-mid)*tag[x];tag[ch[x][1]]+=tag[x];
28         tag[x]=0;
29     }
30 }
31 
32 void Update(int &x,int l,int r,int a,int b){
33     if(!x)x=++tot;
34     Push_down(x,l,r);
35     if(l>=a&&r<=b){sum[x]+=r-l+1;tag[x]+=1;return;}
36     int mid=(l+r)>>1;
37     if(mid>=a)Update(ch[x][0],l,mid,a,b);
38     if(mid<b)Update(ch[x][1],mid+1,r,a,b);
39     Push_up(x);
40 }
41 
42 void Modify(int x,int l,int r,int g,int a,int b){
43     Update(rt[x],1,n,a,b);
44     if(l==r)return;
45     int mid=(l+r)>>1;
46     if(mid>=g)Modify(x<<1,l,mid,g,a,b);
47     else Modify(x<<1|1,mid+1,r,g,a,b);
48 }
49 
50 int Query(int x,int l,int r,int a,int b){
51     if(!x)return 0;
52     Push_down(x,l,r);
53     if(l>=a&&r<=b)return sum[x];
54     int mid=(l+r)>>1,ret=0;
55     if(mid>=a)ret=Query(ch[x][0],l,mid,a,b);
56     if(mid<b)ret+=Query(ch[x][1],mid+1,r,a,b);
57     return ret;
58 }
59 
60 int Solve(int l,int r,int k){
61     int lo=1,hi=n,p=1;
62     while(lo<hi){
63         int mid=(lo+hi)>>1,tmp;
64         tmp=Query(rt[p<<1],1,n,l,r);
65         if(tmp>=k)hi=mid,p<<=1;
66         else lo=mid+1,p=p<<1|1,k-=tmp;
67     }
68     return hi;
69 }
70 
71 int op,a,b,c;
72 int main(){
73 #ifndef ONLINE_JUDGE
74     freopen("zjoi13_sequence.in","r",stdin);
75     freopen("zjoi13_sequence.out","w",stdout);
76 #endif    
77     scanf("%d%d",&n,&m);
78     while(m--){
79         scanf("%d%d%d%d",&op,&a,&b,&c);
80         if(op==1)Modify(1,1,n,n-c+1,a,b);
81         else printf("%d\n",n-Solve(a,b,c)+1);
82     }
83     //printf("%.2f\n",(double)clock()/CLOCKS_PER_SEC);
84     return 0;
85 }