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
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
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]);
}