DP-POJ-3616-Milking Time

Milking Time
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6318 Accepted: 2659
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_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

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：
有M个时间区间存在于0~N这个时间段中，每个时间区间有一个开始时间、结束时间和区间效率（其实是价值），要求每选择一个时间区间，下一个一定要间隔不小于R，求能够积累的最大价值。
一开始没看清题意，写了个value=ef*time，算区间总价值，WA到哭，读完题意一下就过了。

//
// main.cpp
// 基础DP1-R-Milking Time
//
// Created by 袁子涵 on 15/10/28.
// Copyright © 2015年 袁子涵. All rights reserved.
//
// 16ms 748KB

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <stdlib.h>
#define MAX 1005

using namespace std;

typedef struct {
    long long int start,end,ef;
}interval;

long long int N,M,R;
interval inter[MAX];
long long int dp[MAX];

int comp(interval a,interval b)
{
    return a.start<b.start;
}

long long int DP(long long int num)
{
    for (long long int i=1; i<=num-1; i++)
        if (inter[i].end+R<=inter[num].start)
            dp[num]=max(dp[num],dp[i]+inter[num].ef);
    return dp[num];
}

int main(int argc, const char * argv[]) {
    long long int out=0;
    scanf("%lld%lld%lld",&N,&M,&R);
    for (long long int i=1; i<=M; i++) {
        scanf("%lld%lld%lld",&inter[i].start,&inter[i].end,&inter[i].ef);
    }
    sort(inter+1, inter+1+M, comp);
    for (long long int i=1; i<=M; i++)
        dp[i]=inter[i].ef;
    for (long long int i=1; i<=M; i++) {
        out=max(out,DP(i));
    }
    printf("%lld\n",out);
    return 0;
}