HDU 5124 lines

lines

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 515    Accepted Submission(s): 241

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

 

BC#20的1002题

区间的数据范围去到1e8

只要离散一下再用差分标记搞就可以了

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <algorithm>

using namespace std;

typedef long long LL;
typedef pair<int,int> pii;
#define X first
#define Y second
const int N = 300010 ;
const int inf = 1e9+7;
int cnt[N] , tot ;
struct node {
    int x , xx , id ;
}e[N];

bool cmp1( const node &a , const node &b ){ return a.x < b.x ; }
bool cmp2( const node &a , const node &b ){ return a.id < b.id ; }

void run() {

    memset( cnt , 0 , sizeof cnt ) ;
    int n ; tot = 0 ;
    scanf("%d",&n);
    for( int i = 0 ; i < 2 * n ; ++i ){
        scanf("%d",&e[i].x);
        e[i].id = i ;
    }
    sort( e , e + 2 * n , cmp1 );
    e[0].xx = 1 ;
    for( int i = 1 ; i < 2 * n ; ++i ){
        e[i].xx = ( e[i].x == e[i-1].x ? e[i-1].xx : e[i-1].xx + 1 );
    }
    sort( e, e + 2 * n , cmp2 ) ;

    for( int i = 0 ; i < 2 * n ; i += 2 ){
        cnt[ e[i].xx ] ++;
        cnt[ e[i+1].xx + 1 ] --;
    }
    int ans = 0 , sum = 0 ;
    for( int i = 0 ; i <= 2 * n ; ++i ){
        sum += cnt[i];
        ans = max( ans , sum ) ;
    }
    printf("%d\n",ans);
}

int main()
{
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    ios::sync_with_stdio(false);
    int _ ; scanf("%d",&_);
    while( _-- ) run() ;
}