lines-HDU5124（区间处理 +离散化）

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

 

 

题目大意：

给你几个区间，然后给这区间间的点染色

求最染色最多的点 染了多少色

 

分析：

 

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<vector>
using namespace std;
#define N 1100000

struct node
{
    int x,y;
}a[N];
int f[N],p[N];

int main()
{
    int T,n,i;
    scanf("%d",&T);
    while(T--)
    {
        int k=0;
        memset(a,0,sizeof(a));
        memset(f,0,sizeof(f));
        memset(p,0,sizeof(p));
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d %d",&a[i].x,&a[i].y);
            p[k++]=a[i].x;
            p[k++]=a[i].y;
        }
        sort(p,p+k);
        int len=unique(p,p+k)-p;
        int Max=0;
        for(i=0;i<n;i++)
        {
            int u=lower_bound(p,p+len,a[i].x)-p;
            int v=lower_bound(p,p+len,a[i].y)-p;
            f[u]++;
            f[v+1]--;
            Max=max(Max,v+1);
        }
        int ans=0,b=0;
        for(i=0;i<=Max;i++)
        {
            ans+=f[i];
            b=max(b,ans);
        }
        printf("%d\n",b);
    }
    return 0;
}