bzoj 1664 （贪心）

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

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 487  Solved: 344
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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

 

这里很容易贪心证明，不需要去O（n^2）去跑，很容易被卡。

就是这里的，想象成许多线段，如果相交，若当前线段右端比以前线段右端小，那么就赋值为现在那条线段，因为这样只会更优。

如果不想交，直接ans+1

划水

 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<iostream>
 #include<cstdio>

 #define N 10007
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while(ch<=''&&ch>='')
     {
         x=(x<<)+(x<<)+ch-'';
         ch=getchar();
     }
     return x*f;
 }

 int n,ans;
 struct Node
 {
     int x,y;
 }a[N];

 bool cmp(Node x,Node y)
 {
     return x.x<y.x;
 }
 int main()
 {
     n=read();
     for (int i=;i<=n;a[i].x=read(),a[i].y=a[i].x+read()-,i++);
     sort(a+,a+n+,cmp);

     for (int i=,r=;i<=n;i++)
     {
         if (a[i].x>r) ans++,r=a[i].y;
         else if (a[i].y<r) r=a[i].y;
     }
     printf("%d",ans);
 }