hdu 5124 lines（树状数组）_hdu - 5124 lines树状-优快云博客

本文链接：https://blog.youkuaiyun.com/acm_cxq/article/details/51893269

本文介绍了一种解决区间覆盖问题的方法，通过使用树状数组实现区间更新和单点查询，找到被最多线段覆盖的点及其覆盖次数。文章还讨论了如何进行离散化处理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

lines

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1575 Accepted Submission(s): 656

Problem Description

John has several lines. The lines are covered on the X axis. Let A is a point which is covered by the most lines. John wants to know how many lines cover A.

Input

The first line contains a single integer

T(1≤T≤100)(the data for

N>100 less than 11 cases),indicating the number of test cases.
Each test case begins with an integer

N(1≤N≤105),indicating the number of lines.
Next N lines contains two integers

Xi and

Yi(1≤Xi≤Yi≤109),describing a line.

Output

For each case, output an integer means how many lines cover A.

Sample Input

Sample Output

3
1

题意：x轴上很多段区间，求覆盖最多的一个点的覆盖次数

思路：首先很容易想到这个点一定可以是某一段区间的端点，然后这里给了区间，要求一个点的出现次数，想到什么？

没错，区间更新，单点求和。当然用线段树更加直观，只是本人确实不是很喜欢线段树...于是就用树状数组实现了

与单点求和，区间求和的区别在于更新区间的时候add(l,1)和add(r+1,-1) 单点求和就是直接sum(i)就可以了~

关于为什么是这样，本人也没怎么理解深刻，就不多解释了（快上车）。

注意本题要离散化处理

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 100500
int a[N],b[N],p[2*N];
int c[N*2];
int maxn;
int lowbit(int x)
{
    return x&-x;
}
void add(int x,int v)
{
    for(int i=x; i<=maxn; i+=lowbit(i))
        c[i]+=v;
}
int sum(int x)
{
    int ans=0;
    for(int i=x; i; i-=lowbit(i))
        ans+=c[i];
    return ans;
}
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        memset(c,0,sizeof(c));
        scanf("%d",&n);
        maxn=2*n;
        for(int i=1; i<=n; i++)
        {
            scanf("%d %d",&a[i],&b[i]);
            p[i]=a[i],p[i+n]=b[i];
        }
        sort(p+1,p+1+2*n);
        for(int i=1; i<=n; i++)
        {
            int l=lower_bound(p+1,p+1+2*n,a[i])-p;
            int r=lower_bound(p+1,p+1+2*n,b[i])-p;
            add(l,1);
            add(r+1,-1);
        }
        int ans=1;
        for(int i=1;i<=2*n;i++)
        {
            int s=sum(i);
            ans=max(ans,s);
        }
        printf("%d\n",ans);
    }
    return 0;
}