bzoj1664：参加节日庆祝

1664: [Usaco2006 Open]County Fair Events 参加节日庆祝

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 286  Solved: 207
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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

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

4

OUTPUT DETAILS:

FJ can do no better than to attend events 1, 2, 3, and 4.

HINT

 

Source

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 ",f[n]);
    return 0;
}

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