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;
}