hdu5124——lines

最新推荐文章于 2019-10-26 16:15:14 发布

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本文探讨了一种使用线性时间复杂度的算法解决覆盖问题的方法，通过离散化和树状数组优化，实现了高效的计算。详细介绍了输入格式、问题描述、解题思路以及代码实现，适合寻求算法优化的读者。

lines

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 330 Accepted Submission(s): 155

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

Sample Output

3
1

Source

BestCoder Round #20

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Statistic | Submit | Discuss | Note

比赛时一开始用线段树WA了，然后树状数组过的，赛后听某牛说可以O(n)，然后仔细想了下，比如现在要在[l，r]上覆盖一次，那么我们可以在s[l]上+1，s[r + 1] 上 - 1，记录覆盖次数，然后for一遍，当前的s[i] 加上 s[i - 1]就是i这个点被覆盖的次数，因为有s[r + 1] --，所以此方案是合理的，数据太大可以离散化一下

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

using namespace std;

const int N = 100010;

int xis[N << 1];
int po[N << 1];

struct node
{
	int l, r;
}lines[N];

int main()
{
	int t, n, cnt, res, ans;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		map <int, int> newx;
		memset (xis, 0, sizeof(xis));
		newx.clear();
		cnt = 0;
		res = 0;
		for (int i = 0; i < n; ++i)
	 	{
		 	scanf("%d%d", &lines[i].l, &lines[i].r);
		 	po[cnt++] = lines[i].l;
		 	po[cnt++] = lines[i].r;
	 	} 
	 	sort (po, po + cnt);
	 	for (int i = 0; i < cnt; ++i)
	 	{
	 		if (i == 0)
	 		{
	 			newx[po[i]] = ++res;
	 		}
	 		else if(po[i] == po[i - 1])
	 		{
	 			newx[po[i]] = newx[po[i - 1]];
	 		}
	 		else
	 		{
	 			newx[po[i]] = ++res;
	 		}
	 	}
	 	for (int i = 0; i < n; ++i)
	 	{
	 		int l = newx[lines[i].l];
	 		int r = newx[lines[i].r];
	 		xis[l]++;
	 		xis[r + 1]--;
	 	}
	 	ans = 0;
	 	for (int i = 1; i <= res; ++i)
	 	{
	 		xis[i] += xis[i - 1];
	 		if (ans < xis[i])
	 		{
	 			ans = xis[i];
	 		}
	 	}
	 	printf("%d\n", ans);
	}
	return 0;
}