hdu 5124 lines（树状数组）

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

2
5
1 2
2 2
2 4
3 4
5 1000
5
1 1
2 2
3 3
4 4
5 5

 

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
,b
,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;
}