bzoj1664 [Usaco2006 Open]County Fair Events 参加节日庆祝

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.

我会n^2的算法耶……幸好数据弱

首先把每个事件的开始时间、结束时间提出来快排，然后令f[i]表示快排后前i个最多能取多少个，枚举如果f[j].t<f[i].s，那么事件j一定在i前面，就可以用j来更新答案

其实注意到if (e[j].t<e[i].s) f[i]=max(f[i],f[j]+1)这一行，显然可以用平衡树加速，但是我很懒，又不会STL的set，就不打了

#include<cstdio>

#include<algorithm>

using namespace std;

struct event{

	int s,t;

}e[10010];

int n;

int f[10010];

inline bool cmp(const event &a,const event &b)

{return a.s<b.s||a.s==b.s&&a.t<b.t;}

inline int max(int a,int b)

{return a>b?a:b;}

inline int read()

{

    int x=0,f=1;char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

int main()

{

	n=read();

	for (int i=1;i<=n;i++)

	  {

	  	e[i].s=read();

	  	e[i].t=e[i].s+read()-1;

	  }

	sort(e+1,e+n+1,cmp);

	for(int i=1;i<=n;i++)

	  {

	  	f[i]=1;

	  	for (int j=1;j<i;j++)

	  	  if (e[j].t<e[i].s) f[i]=max(f[i],f[j]+1);

	  }

	printf("%d",f[n]);

}

巴特西