51nod 1428 && bzoj 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

Time Limit: 10 Sec   Memory Limit: 64 MB
Submit: 951   Solved: 550
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Description

有N头牛,每头牛有个喝水时间,这段时间它将专用一个Stall 现在给出每头牛的喝水时间段,问至少要多少个Stall才能满足它们的要求

Input

* Line 1: A single integer, N

* Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.

Output

* Line 1: The minimum number of stalls the barn must have.

* Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.

Sample Input

5

1 10

2 4

3 6

5 8

4 7

Sample Output

4

经典题了

把线段转化成一个入点和一个出点，这样存下2*n个点，之后按坐标从小到大给这2*n个点排序，之后O(n)遍历一遍就好，遇到入点sum++, 遇到出点sum--，遍历时最大的sum便是答案，算上排序复杂度稳定nlogn（注意：排序时坐标相同的点出点优先，出点不在区间内）

#include<stdio.h>
#include<algorithm>
using namespace std;
typedef struct
{
	int x;
	int flag;
}Point;
Point s[100005];
bool comp(Point a, Point b)
{
	if(a.x<b.x || a.x==b.x && a.flag<b.flag)
		return 1;
	return 0;
}
int main(void)
{
	int n, i, a, b, sum, ans;
	scanf("%d", &n);
	for(i=1;i<=n;i++)
	{
		scanf("%d%d", &a, &b);
		s[i*2-1].x = a, s[i*2-1].flag = 1;
		s[i*2].x = b+1, s[i*2].flag = -1;
	}
	n *= 2;
	sort(s+1, s+n+1, comp);
	sum = ans = 0;
	for(i=1;i<=n;i++)
	{
		sum += s[i].flag;
		ans = max(ans, sum);
	}
	printf("%d\n", ans);
	return 0;
}