[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

Sample Output

4

---

首先按结束时间排序，然后一路搜下去，发现一个节日的开始时间在所记录的结束时间之后，就参加这个节日，同时更新记录的结束时间（典型贪心思想）

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
using namespace std;
inline int read(){
	int x=0,f=1;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())  if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())   x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x>=10) print(x/10);
	putchar(x%10+'0');
}
const int N=1e4;
struct AC{
	int l,r;
	void join(int x,int y){l=x,r=y;}
	bool operator <(const AC &x)const{return r<x.r;}
}A[N+10];
int main(){
	int n=read();
	for (int i=1,x,y;i<=n;i++)   x=read(),y=read(),A[i].join(x,x+y);
	sort(A+1,A+1+n);
	int x=A[1].r,ans=1;
	for (int i=2;i<=n;i++)   if (A[i].l>=x)   x=A[i].r,ans++;
	printf("%d\n",ans);
	return 0;
}