HDUOJ lines

/*


lines
Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 590    Accepted Submission(s): 277



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


 


Source

 BestCoder Round #20  
*/

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int comp(const void *a,const void *b)
{
	return *(int *)a-*(int *)b;
}
int a[100005],b[100005];
int main()
{
	int cases,j,i,temp_j,temp_i,p,cnt,max,n;
	scanf("%d",&cases);
	while(cases--)
	{
		memset(a,-1,sizeof(a));
		memset(b,-1,sizeof(b));
		max=-1,cnt=0;
		scanf("%d",&n);
		for(i=0;i<n;i++)
		{
			scanf("%d%d",&a[i],&b[i]);
		}
		qsort(a,100005,sizeof(int),comp);
		qsort(b,100005,sizeof(int),comp);
		for(j=0;;j++){if(b[j]!=-1) break;}
		temp_j=j;
		for(i=0;i<100005;i++)
		{

			if(a[i]==-1) continue;
			if(a[i]>b[j])
            {
                cnt--;
                j++;
                temp_j=j;
                break;
            }

			else
			{
			    p=1;
				cnt++;temp_i=i;
				if(cnt>max) max=cnt;
				while(p)
				{
					if(a[++temp_i]==a[i]){break;}
					while(1)
					{
						if(b[temp_j++]==a[i]) {cnt--;j++;temp_j=j;}
						else {p=0;temp_j=j;break;}
					}
				}
			}
		}
		printf("%d\n",max);

	}
	return 0;
}

//这题是一次bc的b题，下面是大仙的代码，和我的想法一样， 不过他用结构体标记左右端点，再混合排序，而我对结构体不熟悉
//没想到用这个所以就用a，b数组讨论成数组的形式，思维情况比较复杂，wa了5，6次。最后还是ac了 ，时间和大仙的那个差不多，差了100ms左右。 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct line{
    int x;
    int flag;
}L[200010];
bool cmp(line a,line b){
    if(a.x<b.x||(a.x==b.x&&a.flag<b.flag)) return 1;
    return 0;
}
int main()
{
    int _,n,all,a,b;
    scanf("%d",&_);              //用_作为循环变量，值得学习。 
    while(_--){
        scanf("%d",&n);
        all=0;
        for(int i=1;i<=n;i++){
            scanf("%d%d",&a,&b);
            L[all].x=a;
            L[all++].flag=0;
            L[all].x=b;
            L[all++].flag=1;
        }
        sort(L,L+all,cmp);
        int total=0;
        int ma=0;
        for(int i=0;i<all;i++){
            if(L[i].flag==0)
                total++;
            else total--;
            if(total>ma)
                ma=total;
        }
        printf("%d\n",ma);
    }
    return 0;
}
//大仙的ac的代码。