题
Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her next N (1 ≤ N ≤ 1,000,000) hours (conveniently labeled 0…N-1) so that she produces as much milk as possible.
Farmer John has a list of M (1 ≤ M ≤ 1,000) possibly overlapping intervals in which he is available for milking. Each interval i has a starting hour (0 ≤ starting_houri ≤ N), an ending hour (starting_houri < ending_houri ≤ N), and a corresponding efficiency (1 ≤ efficiencyi ≤ 1,000,000) which indicates how many gallons of milk that he can get out of Bessie in that interval. Farmer John starts and stops milking at the beginning of the starting hour and ending hour, respectively. When being milked, Bessie must be milked through an entire interval.
Even Bessie has her limitations, though. After being milked during any interval, she must rest R (1 ≤ R ≤ N) hours before she can start milking again. Given Farmer Johns list of intervals, determine the maximum amount of milk that Bessie can produce in the N hours.
Input
- Line 1: Three space-separated integers: N, M, and R
- Lines 2…M+1: Line i+1 describes FJ’s ith milking interval withthree space-separated integers: starting_houri , ending_houri , and efficiencyi
Output
- Line 1: The maximum number of gallons of milk that Bessie can product in the N hours
-
翻译
- 贝西在时间内可以产奶
- 农夫约翰有一些时间间隔（1≤ M≤ 1000），在时间间隔内他可以帮助贝西产奶。每个间隔都有一个起始小时（0≤ 开始时间≤ N） ，一个结束的小时（开始时间<结束时间≤ N）
- 农夫约翰分别在开始时间和结束时间开始和停止挤奶。在挤奶时，贝西必须在整个时间间隔内产奶。贝西也有她的局限性。在产奶结束后，她必须休息R（1≤ R≤ N） 几小时后她才能开始挤奶。
- 求贝西在N小时内可以生产的最大牛奶量。
Sample Input
12 4 2
1 2 8
10 12 19
3 6 24
7 10 31
Sample Output
43
代码
#include "stdio.h"
#include "algorithm"
#include "string.h"
using namespace std;
int dp[1009];
struct ppp
{
int a,b,c;
}p[1009],h;
int qqq(ppp x,ppp y)
{
if(x.a==y.a)
return x.b<y.b;
return x.a<y.a;
}
int main()
{
int i,m,n,j,k,l=0,s;
scanf("%d%d%d",&s,&n,&k);
for(i=0;i<n;i++)
{
scanf("%d %d %d",&p[i].a,&p[i].b,&p[i].c);
p[i].b+=k;
}
sort(p,p+n,qqq);
for(i=0;i<n;i++)
{
//因为用时间段来写dp数组很麻烦，所以我们就用它的排列顺序来写dp数组
dp[i]=p[i].c;//我们是以它的排列顺序写的dp数组,和平时有些不一样
for(j=0;j<i;j++)
{
if(p[j].b<=p[i].a)
{
dp[i]=max(dp[i],dp[j]+p[i].c);
l=max(l,dp[i]);
}
}
}
printf("%d\n",l);
}