bzoj 1664: [Usaco2006 Open]County Fair Events 参加节日庆祝（DP）

最新推荐文章于 2018-12-11 21:37:00 发布

Jaihk662

最新推荐文章于 2018-12-11 21:37:00 发布

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本文介绍了一种解决农民约翰如何参加最多节日庆祝的问题的方法。通过将节日按结束时间排序并使用动态规划算法来确定最大参与数量。示例输入输出展示了算法的有效性。

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1664: [Usaco2006 Open]County Fair Events 参加节日庆祝

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 456 Solved: 321
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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

1 6

8 6

14 5

19 2

1 8

18 3

10 6

Sample Output

貌似像这种线段按尾排序然后DP的题已经好多道了（第7页的）

dp[i]表示前i小时能参加的最多活动数量

#include<stdio.h>
#include<algorithm>
using namespace std;
typedef struct Line
{
	int x, y;
	bool operator < (const Line &b) const
	{
		if(y<b.y)
			return 1;
		return 0;
	}
}Line;
Line s[10005];
int dp[200005];
int main(void)
{
	int n, i, p;
	scanf("%d", &n);
	for(i=1;i<=n;i++)
	{
		scanf("%d%d", &s[i].x, &s[i].y);
		s[i].y += s[i].x-1;
	}
	sort(s+1, s+n+1);
	p = 1;
	for(i=1;i<=200000;i++)
	{
		dp[i] = dp[i-1];
		while(s[p].y==i)
		{
			dp[i] = max(dp[i], dp[s[p].x-1]+1);
			p++;
		}
	}
	printf("%d\n", dp[200000]);
	return 0;
}