poj 3616(动态规划)

Milking Time

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

Source

USACO 2007 November Silver

题意：挤奶工工作的区间是 1 - N (hour),农场主有m个时间段要挤奶工挤奶,每一个时间段都有一个 起始时间 结束时间 以及在这段时间内可以获得的利益 。而且挤奶工每工作

完一个时间段就要休息r分钟，求挤奶工在 这m个时间段中能够获得的最大利益。

题解：dp[i]表示在前 i 段时间挤奶工能够获得的最大收益,对前m段时间按起始时间升序排个序，起始时间相同按照结束时间升序排列。我们就可以得到状态转移方程. dp[i] = max(dp[j]+value[i],dp[i])&&mk[j].e+r<=mk[i].s (1=<j<i)

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int N = 1000005;
const int M = 1005;
int dp[N]; ///dp[i]表示在前i组时间中能取得的最大利益
struct Milk{
    int s,e,v;
}mk[M];
int cmp(Milk a,Milk b){
    if(a.s!=b.s) return a.s<b.s;
    return a.e<b.e;
}
int main()
{
    int n,m,r;
    while(scanf("%d%d%d",&n,&m,&r)!=EOF)
    {
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&mk[i].s,&mk[i].e,&mk[i].v);
        }
        memset(dp,0,sizeof(dp));
        sort(mk+1,mk+1+m,cmp);
        int mx = -1;
        for(int i=1;i<=m;i++){
            dp[i] = mk[i].v;
            for(int j=1;j<i;j++){
                if(mk[j].e+r<=mk[i].s&&mk[i].v+dp[j]>dp[i]){
                    dp[i] = dp[j] +mk[i].v;
                }
            }
            if(dp[i]>mx) mx = dp[i];
        }
        printf("%d\n",mx);
    }
    return 0;
}