BZOJ 1664 [Usaco2006 Open]County Fair Events 参加节日庆祝贪心

最新推荐文章于 2020-05-05 21:19:26 发布

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最新推荐文章于 2020-05-05 21:19:26 发布

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分类专栏：贪心

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本文介绍了一个经典的时间安排问题：如何最大化参与活动的数量。给定一系列活动的开始时间和结束时间，目标是找出能参加的最大数量的非重叠活动。文章提供了一个有效的解决方案，并附带了解题思路和实现代码。

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Description

Farmer John has returned to the County Fair so he can attend the special events (concerts, rodeos, cooking shows, etc.). He wants to attend as many of the N (1 <= N <= 10,000) special events as he possibly can. He's rented a bicycle so he can speed from one event to the next in absolutely no time at all (0 time units to go from one event to the next!). Given a list of the events that FJ might wish to attend, with their start times (1 <= T <= 100,000) and their durations (1 <= L <= 100,000), determine the maximum number of events that FJ can attend. FJ never leaves an event early.

有N个节日每个节日有个开始时间,及持续时间. 牛想尽可能多的参加节日,问最多可以参加多少. 注意牛的转移速度是极快的,不花时间.

Input

* Line 1: A single integer, N.

* Lines 2..N+1: Each line contains two space-separated integers, T and L, that describe an event that FJ might attend.

Output

* Line 1: A single integer that is the maximum number of events FJ can attend.

Sample Input

7
1 6
8 6
14 5
19 2
1 8
18 3
10 6

INPUT DETAILS:

Graphic picture of the schedule:
11111111112
12345678901234567890---------这个是时间轴.
--------------------
111111 2222223333344
55555555 777777 666

这个图中1代表第一个节日从1开始,持续6个时间,直到6.

Sample Output

4

OUTPUT DETAILS:

FJ can do no better than to attend events 1, 2, 3, and 4.

HINT

传送门

同bzoj3410……

其实直接转化为区间，

然后就是求最大选几个不重叠区间了……

贴代码的时候竟然忘了这题略有不同……区间是严格不能重叠的。

#include<bits/stdc++.h>
using namespace std;
const int 
	N=10005;
int n;
struct node{int l,r;}a[N];
bool cmp(node x,node y){return x.r<y.r;}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
		scanf("%d%d",&a[i].l,&a[i].r),a[i].r+=a[i].l-1;
	sort(a+1,a+1+n,cmp);
	int r=a[1].r,ans=1;
	for (int i=2;i<=n;i++)
		if (a[i].l>r) r=a[i].r,ans++;
	printf("%d\n",ans);
	return 0;
}