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個節日每個節日有個開始時間,及持續時間. 牛想盡可能多的參加節日,問最多可以參加多少. 註意牛的轉移速度是極快的,不花時間.

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

首先按結束時間排序，然後一路搜下去，發現一個節日的開始時間在所記錄的結束時間之後，就參加這個節日，同時更新記錄的結束時間（典型貪心思想）

#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;
}
 

[Usaco2006 Open]County Fair Events 參加節日慶祝