POJ 3616 Milking Time 基础DP

　　　　　　　　　　　　　　　　　　　　　　Milking Time

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5743		Accepted: 2401

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：有一只牛，他在0~N的时间内可以产牛奶，现在有M个区间，你可以指定这M个区间的任意个区间让他去产牛奶，给出这M个区间的开始时间，结束时间，还有在这个区间的时间可以产的牛奶的量。

有个条件，就是你给他指定的产牛奶的区间每2个之间至少要有R的时间让他休息。

现在问怎么指定区间让他去产牛奶，可以获得最大的牛奶。

输出最大的牛奶。

 

基础DP

本质：最长上升子序列。

 

先把区间排序

设dp[i]表示以第i个区间作为让他产牛奶的最后一个区间，可以获得的最大的牛奶量。

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxM=1005;

long long dp[maxM];
struct Node
{
    int u,v;
    long long w;
}node[maxM];

bool cmp(Node a,Node b)
{
    if(a.u==b.u)
        return a.v<b.v;
    else
        return a.u<b.u;
}

int main()
{
    int T,M,R;
    while(~scanf("%d%d%d",&T,&M,&R))
    {
        for(int i=0;i<M;i++)
        {
            scanf("%d%d%lld",&node[i].u,&node[i].v,&node[i].w);
        }
        sort(node,node+M,cmp);

        for(int i=0;i<M;i++)
        {
            if(node[i].u>T||node[i].v>T)
                break;
            dp[i]=node[i].w;
            for(int j=0;j<i;j++)
            {
                if(node[j].v+R<=node[i].u&&dp[j]+node[i].w>dp[i])
                    dp[i]=dp[j]+node[i].w;
            }
        }

        long long ans=-1;
        for(int i=0;i<M;i++)
            if(dp[i]>ans)
                ans=dp[i];
        printf("%lld\n",ans);
    }
    return 0;
}