HDU 5124 lines 最多区间覆盖

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Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 453    Accepted Submission(s): 220

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

251 2 2 22 43 45 100051 12 23 34 45 5

 

Sample Output

31

 

Source

BestCoder Round #20

 

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。

//953MS1792K#include<stdio.h>#include<algorithm>using namespace std;pair<int,int>p[200007];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,a,b;        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d%d",&a,&b);            p[i*2]=make_pair(a,1);            p[i*2+1]=make_pair(b+1,-1);        }        sort(p,p+2*n);//对坐标进行排序        int ans=0,maxx=0;        for(int i=0;i<2*n;i++)        {            ans+=p[i].second;            if(ans>maxx)maxx=ans;        }        printf("%d\n",maxx);    }    return 0;}