[ZJOI 2013] bzoj3110 K大数查询 【树套树】

#include <cstdio>  
  
const int Maxn = 50000 + 10,MS = Maxn << 8;  
int root[Maxn << 2],ls[MS],rs[MS],tot,M;  
long long Cnt[MS],ly[MS];  
//线段树的外层为权值，内层为区间  
  
int Ina; char Inc; bool InSign;  

inline int geti() 
{  
    InSign = 0;  
    while (Inc = getchar(),Inc < '0' || Inc > '9') InSign |= Inc == '-';  
    Ina = Inc - '0';  
    while (Inc = getchar(),Inc >= '0' && Inc <= '9') Ina = (Ina << 3) + (Ina << 1) + Inc - '0';  
    if (InSign) Ina = -Ina;  
    return Ina;  
}  
  
void Insert(int &u,int x,int y) {  //内层线段树 
    if (!u) u = ++tot; 
    if (l >= x && r <= y) {  
        Cnt[u] += r - l + 1;  
		 ++ly[u];//永久化懒惰标记 
		 return;  
    }  
    int mid = (l + r) >> 1;  
    if (y <= mid) Insert(ls[u],x,y);  
    else if (x > mid) Insert(rs[u],mid + 1,y);  
    else {  
        Insert(ls[u],mid);  
        Insert(rs[u],y);  
    }  
    Cnt[u] += y - x + 1;  
}
  
void add(int x,int y,int val) {  //外层线段树 权值线段树 
    int l = 1,r = N,u = 1,mid;  
    while (true) {  
        Insert(root[u],y);//  
        if (l == r) break;  //边界 
        mid = (l + r) >> 1; u <<= 1;  
        if (val <= mid) r = mid;  
        else l = mid + 1,u |= 1;  
    }  
}  
  
long long Count(int u,int y) {  
    if (l >= x && r <= y) return Cnt[u];  
    int mid = (l + r) >> 1; long long ans = 0;  
    if (y <= mid) ans = Count(ls[u],y);  
    else if (x > mid) ans = Count(rs[u],y);  
    else ans = Count(ls[u],mid) + Count(rs[u],y);  
    ans += (y - x + 1) * ly[u];  
    return ans;  
}  
  
int query(int x,long long val) {  
    int l = 1,u = 1; long long tmp;  
    while (true) {  
        if (l == r) return l;  
        mid = (l + r) >> 1;  
        tmp = Count(root[u << 1 | 1],y);  
        if (tmp >= val) l = mid + 1,u = u << 1 | 1;  
        else r = mid,u = u << 1,val -= tmp;  
    }  
}  
  
int main() {  
    N = geti(),M = geti();  
    int type,b; long long c;  
    while (M--) {  
        type = geti(),a = geti(),b = geti(),c = geti();  
        if (type < 2) add(a,c);  
        else printf("%dn",query(a,c));
    }  
    return 0;  
}  

好像还有其他做法。。暂时先贴这个。。什么CDQ分治整体二分我有空再去看qwq

我的代码比较丑=。=

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
int Ina; char Inc; bool InSign;  
inline int read() 
{  
    InSign = 0;  
    while (Inc = getchar(),Inc >= '0' && Inc <= '9') Ina = (Ina << 3) + (Ina << 1) + Inc - '0';  
    if (InSign) Ina = -Ina;  
    return Ina;  
} 
int n,m;
const int Maxn = 50000 + 10,tot;  
long long cnt[MS],ly[MS];  
//线段树的外层为权值，内层为区间  
void insert(int& u,int y)
{
	if(!u)u = ++tot;
	if(l>=x&&r<=y){	cnt[u] += r-l+1;++ly[u];return;}
	int mid = (l+r)>>1;
	if(y<=mid)insert(ls[u],y);  
	else if(x>mid)insert(rs[u],y);
	else
	{
		insert(ls[u],mid);
		insert(rs[u],y);
	}
	cnt[u] += y - x + 1;
}
/*
void add(int x,LL v)
{
	insert(root[u] ? root[u] : root[u] = ++size,n,R);
	if(l==r) return;
	int mid = (l+r)>>1;
	if(x<=mid)add(u<<1,x);
	else add(u<<1|1,x);
}
*/
void add(int x,int v)
{
	int l=1,r=n,u=1,mid;
	while(true)
	{
		insert(root[u],y);
		if(l==r)break;
		mid = (l+r)>>1;u<<=1;
		if(v<=mid)r = mid;
		else l = mid+1,u|=1;
	}
}
LL Count(int u,int y)
{
	if(l>=x&&r<=y)return cnt[u];
	int mid = (l+r)>>1;LL ans = 0;
	if(y<=mid)ans = Count(ls[u],y);
	else if(x>mid)ans = Count(rs[u],y);
	else ans = Count(ls[u],y);
	ans += (y-x+1)*ly[u];//懒惰标记永久化了
	return ans; 
}
/*
递归版：
int query(int u,int x)  
{ // 外层线段树   
    if(l == r) return l;  
      
    int mid = (l + r) >> 1;  
      
    int cnt = count(root[rch],R);  
    if(cnt >= x) return query(rch,x);  
    else return query(lch,x - cnt);  
}   
*/
int query(int x,LL v)
{
	int l = 1,r = n,u = 1;LL tmp;
	while(true)	
	{
		if(l==r)return l;
		mid = (l+r)>>1;
		tmp = Count(root[u<<1|1],y);
		if(tmp>=v)l=mid+1,u=u<<1|1;
		else r=mid,u=u<<1,v-=tmp;
	}
}
int main(void)
{
	n = read(),m = read();
	int type,b;LL c;
	for(int i=1;i<=m;i++)	
	{
		type = read(),a = read(),b = read(),c = read();
		if(type == 1)add(a,c);
		else printf("%dn",c));
	}
	return 0;
}

STL暴力大法 http://blog.csdn.net/qq_21110267/article/details/44514709

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