BZOJ 1664 [Usaco2006 Open]County Fair Events 参加节日庆祝 贪心
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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
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
OUTPUT DETAILS:
FJ can do no better than to attend events 1, 2, 3, and 4.
HINT
#include<bits/stdc++.h>using namespace std;const int N=10005;int n;struct node{int l,r;}a[N];bool cmp(node x,node y){return x.r<y.r;}int main(){scanf("%d",&n);for (int i=1;i<=n;i++)scanf("%d%d",&a[i].l,&a[i].r),a[i].r+=a[i].l-1;sort(a+1,a+1+n,cmp);int r=a[1].r,ans=1;for (int i=2;i<=n;i++)if (a[i].l>r) r=a[i].r,ans++;printf("%d\n",ans);return 0;}
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