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.
* 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.
* Line 1: A single integer that is the maximum number of events FJ can attend.
Silver
就是先排个序然后就是递推一下吧，好像白书里面有贪心的算法但是翻了很久就是翻不到。。
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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
using namespace std;
const int nmax=10005;
int f[nmax];
struct edge{
int l,r;
bool operator<(const edge&rhs) const {
return l<rhs.l||l==rhs.l&&r<rhs.r;}
};
edge a[nmax];
int main(){
int n;
scanf("%d",&n);
for(int i=1,o;i<=n;i++){
scanf("%d%d",&a[i].l,&o);
a[i].r=a[i].l+o-1;
}
sort(a+1,a+n+1);
for(int i=1;i<=n;i++){
f[i]=1;
for(int j=1;j<i;j++){
if(a[i].l>a[j].r)
f[i]=max(f[i],f[j]+1);
}
}
printf("%d\n",f[n]);
return 0;
}
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