Milking Time---poj3616

Description

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_hour_i ≤ N), an ending hour (starting_hour_i < ending_hour_i ≤ N), and a corresponding efficiency (1 ≤ efficiency_i ≤ 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_hour_i , ending_hour_i , and efficiency_i

Output

* Line 1: The maximum number of gallons of milk that Bessie can product in the N hours

Sample Input

12 4 2
1 2 8
10 12 19
3 6 24
7 10 31

Sample Output

43

题目大意：   
刚开始题目一直都不懂   到底不知道是让干啥呢   更不知道答案是怎么得来的
搜了好几个题解  最后知道了  他是说啥意思吧
就是给你m个挤奶的时间段和挤奶的量
就是这个人挤奶的话    必须是一个时段全都在挤奶   他挤过这段时间他必须休息r小时之后再开始下一次挤
最后求n个小时最大的挤奶量

分析：
就是dp  没啥好说的

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<stdlib.h>

using namespace std;

#define INF 0xfffffff
#define N 1100
struct node
{
    int s,e,l;
}a[N];

int cmp(const void *x,const void *y)
{
    struct node *c,*d;
    c=(struct node *)x;
    d=(struct node *)y;
    if(c->e!=d->e)
        return c->e-d->e;
    else
        return c->s-d->s;
}

int main()
{
    int n,m,r,dp[N];
    while(scanf("%d %d %d",&n,&m,&r)!=EOF)
    {
        for(int i=0;i<m;i++)
        {
            scanf("%d %d %d",&a[i].s,&a[i].e,&a[i].l);
        }
        qsort(a,m,sizeof(a[0]),cmp);
        for(int i=0;i<m;i++)
        {
            dp[i]=a[i].l;
        }
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<i;j++)
            {
                if(a[j].e+r<=a[i].s)
                {
                    dp[i]=max(dp[i],dp[j]+a[i].l);
                }
            }
        }
        int Max=0;
        for(int i=0;i<m;i++)
        {
            Max=max(Max,dp[i]);
        }
        printf("%d\n",Max);
    }
    return 0;
}