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

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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