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.
* 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.
* Line 1: A single integer that is the maximum number of events FJ can attend.
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 }
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