lines

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1350    Accepted Submission(s): 558

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

官方题解：

我们可以将一条线段[xi,yi]分为两个端点xi和(yi)+1，在xi时该点会新加入一条线段，同样的，在(yi)+1时该点会减少一条线段，因此对于2n个端点进行排序，令xi为价值1，yi为价值-1，问题转化成了最大区间和，因为1一定在-1之前，因此问题变成最大前缀和，我们寻找最大值就是答案，另外的，这题可以用离散化后线段树来做。复杂度为排序的复杂度即nlgn，另外如果用第一种做法数组应是2n，而不是n，由于各种非确定性因素我在小数据就已经设了n=10W的点。

题解：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #define mem(x,y) memset(x,y,sizeof(x))
 7 using namespace std;
 8 typedef long long LL;
 9 const int INF=0x3f3f3f3f;
10 const int MAXN=1e5+100;
11 pair<int,int>pa[MAXN<<1];
12 int main(){
13     int T,N;
14     scanf("%d",&T);
15     while(T--){int a,b;
16         scanf("%d",&N);
17         for(int i=0;i<N;i++){
18             scanf("%d%d",&a,&b);
19             pa[i<<1]=make_pair(a,1);
20             pa[i<<1|1]=make_pair(b+1,-1);
21         }
22         sort(pa,pa+N*2);
23         int ans=0,cnt=0;
24         for(int i=0;i<2*N;i++)
25             cnt+=pa[i].second,ans=max(ans,cnt);
26         printf("%d\n",ans);
27     }
28     return 0;
29 }