【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\n",ans[i][1]-ans[i][0]);}}    return 0;}