TOJ 4105 Lines Counting（离线树状数组）

Posted 2020-08-30 Blackops

4105.   Lines Counting

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Time Limit: 2.0 Seconds   Memory Limit: 150000K
Total Runs: 152   Accepted Runs: 47

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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.

Sample Input

3
1 3
2 4
3 5
2
1 2 3 4
1 4 5 6

Sample Output

2
1

 

 

题目链接：TOJ 4105

题意就是在给你N条在X轴上的线段，求左端点在L1~R1且右端点在L2~R2的线段条数，其实这题跟NBUT上一道题很像，问你在区间L1~R1中，值在L2~R2中有几个数，只是这题在起点计数回退时可能多退几个位子，因为线段的起点和终点坐标可能有重复的，都不能算进去，因此要用while语句来操作。离线树状数组是什么个意思呢？就是满足区间减法这样原理的话就可以这样计数：在遇到起点时减去起点到询问区间起点-1的count1，在遇到计数终点时加上询问起点~询问终点的count2，这样可以发现count1其实是count2的子集，一加一减一定会把count1抵消掉，只留下刚好符合询问区间的答案了。

代码：

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N=100010;
struct Line
{
    int l,r;
    bool operator<(const Line &rhs)const
    {
        if(l!=rhs.l)
            return l<rhs.l;
        return r<rhs.r;
    }
};
struct query
{
    int k,r1,r2,flag,id;
    query(int _k=0,int _r1=0,int _r2=0,int _flag=0,int _id=0):k(_k),r1(_r1),r2(_r2),flag(_flag),id(_id){}
    bool operator<(const query &rhs)const
    {
        return k<rhs.k;
    }
};
Line line[N];
query Q[N<<1];
int T[N],ans[N];

void init()
{
    CLR(T,0);
    CLR(ans,0);
}
void add(int k,int v)
{
    while (k<N)
    {
        T[k]+=v;
        k+=(k&-k);
    }
}
int getsum(int k)
{
    int ret=0;
    while (k)
    {
        ret+=T[k];
        k-=(k&-k);
    }
    return ret;
}
int main(void)
{
    int n,m,i;
    while (~scanf("%d",&n))
    {
        init();
        for (i=0; i<n; ++i)
            scanf("%d%d",&line[i].l,&line[i].r);
        sort(line,line+n);
        scanf("%d",&m);
        int qcnt=0;
        for (i=0; i<m; ++i)
        {
            int l1,r1,l2,r2;
            scanf("%d%d%d%d",&l1,&r1,&l2,&r2);
            Q[qcnt++]=query(l1,l2,r2,0,i);
            Q[qcnt++]=query(r1,l2,r2,1,i);
        }
        sort(Q,Q+qcnt);
        int x=0;
        for (i=0; i<qcnt; ++i)
        {
            while (line[x].l<=Q[i].k&&x<n)
                add(line[x++].r,1);
            if(Q[i].flag)
                ans[Q[i].id]+=getsum(Q[i].r2)-getsum(Q[i].r1-1);
            else
            {
                while (line[x-1].l>=Q[i].k&&x-1>=0)
                {
                    add(line[x-1].r,-1);
                    --x;
                }
                ans[Q[i].id]-=getsum(Q[i].r2)-getsum(Q[i].r1-1);
                while (line[x].l<=Q[i].k&&x<n)
                {
                    add(line[x].r,1);
                    ++x;
                }
            }
        }
        for (i=0; i<m; ++i)
            printf("%d\n",ans[i]);
    }
    return 0;
}