【十分不错】【离线+树状数组】【TOJ4105】【Lines Counting】

On the number axis, there are N lines. The two endpoints L and R of each line are integer. Give you M queries, each query contains two intervals: [L1,R1] and [L2,R2], can you count how many lines satisfy this property: L1≤L≤R1 and L2≤R≤R2?

Input

First line will be a positive integer N (1≤N≤100000) indicating the number of lines. Following the coordinates of the N lines' endpoints L and R will be given (1≤L≤R≤100000). Next will be a positive integer M (1≤M≤100000) indicating the number of queries. Following the four numbers L1,R1,L2 and R2 of the M queries will be given (1≤L1≤R1≤L2≤R2≤100000).

Output

For each query output the corresponding answer.

双区间查询

第一个区间就裸的[1...l1-1] [l2..r2] 

   [1.......r1] [l2..r2]

      用vector （Q[i]）维护~(i为l1-1,r1的值)

       表示 [1...i][Q[i][j].l2...Q[i][j].r2] 表示上述~

第二个区间 当枚举i的时候，内循环枚举线段到s[p].x<=i，s[p].y加入a[]来维护终止点

  因为起止点排好序了，

  一定满足Q[i]的[1....i]

  a[]树状数组来维护好l2,r2.

原理理解后 代码很简单。。

直接贴朱神的代码了

复杂度 比较难表述 但是显然是接受范围

（当初考虑l,r不好存。。开数组不好存，才知道可以表示在数组内就好）

#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
#include <algorithm>
#include <vector>
using namespace std;

const int N=100000;

int a[100005];

int inline lowbit(int x){
	return x&(-x);
}
void add(int p,int val){
	while(p<=N){
		a[p]=(a[p]+val);
		p+=lowbit(p);
	}
}
int sum(int p){
	int ans=0;
	while(p>0){
		ans=(ans+a[p]);
		p-=lowbit(p);
	}
	return ans;
}

struct Query{
	int l,r;
	int oriid;
	int orip;
	Query(){}
	Query(int a,int b,int c,int d):l(a),r(b),oriid(c),orip(d){}
};

vector<Query> qs[100005];

int ans[100005][2];

struct Pair{
	int x,y;
	Pair(){}
	Pair(int a,int b):x(a),y(b){}
	bool operator<(const Pair&b)const{
		return x<b.x;
	}
}side[100005];

int main()
{
	int n,m;
	while(~scanf("%d",&n)){
		for(int i=0;i<n;i++){
			scanf("%d%d",&side[i].x,&side[i].y);
		}
		sort(side,side+n);
		for(int i=0;i<=N;i++) qs[i].clear();
		scanf("%d",&m);
		int maxq=0;
		for(int i=0;i<m;i++){
			int a,b,c,d;
			scanf("%d%d%d%d",&a,&b,&c,&d);
			qs[a-1].push_back(Query(c,d,i,0));
			qs[b].push_back(Query(c,d,i,1));
			maxq=max(maxq,(max(a,b)));
		}

		memset(a,0,sizeof(a));
		int sp=0;
		for(int i=0;i<=maxq;i++){
			while(sp<n&&side[sp].x==i){
				add(side[sp].y,1);
				sp++;
			}
			for(int j=0;j<qs[i].size();j++){
				int ans1=sum(qs[i][j].l-1);
				int ans2=sum(qs[i][j].r);
				ans[qs[i][j].oriid][qs[i][j].orip]=ans2-ans1;
			}
		}
		for(int i=0;i<m;i++){
			printf("%d
",ans[i][1]-ans[i][0]);
		}
	}
    return 0;
}