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.
一个普通的dp
按时间轴排序，蓝后
$f[a[i].l+a[i].r-1]=max(f[a[i].l+a[i].r-1],f[a[i].l-1]+1)$
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 int max(int a,int b){return a>b?a:b;} 7 struct data{int l,r;}a[10005]; 8 bool cmp(const data &A,const data &B){return A.r<B.r;} 9 int n,f[100005],m; 10 int main(){ 11 scanf("%d",&n); 12 for(int i=1;i<=n;++i){ 13 scanf("%d%d",&a[i].l,&a[i].r); 14 a[i].r+=a[i].l-1; m=max(m,a[i].r); 15 }sort(a+1,a+n+1,cmp); 16 for(int i=1,j=1;i<=m;++i){ 17 f[i]=max(f[i],f[i-1]); 18 for(;a[j].r<=i&&j<=n;++j) 19 f[i]=max(f[i],f[a[j].l-1]+1); 20 }printf("%d",f[m]); 21 return 0; 22 }