POJ 3616 Milking Time

Language: Default

Milking Time

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13855		Accepted: 5851

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

给定时间段，每个时间段有工作效率，每个时间段都要休息，求最大工作量

跟求最大上升子序列差不多

#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
struct node{
    int s,e,ef;
    
}x[100000];
bool cmp(node a,node b){
    return a.e<b.e;
}
int dp[100000];
int main()
{
    int a,b,c;
    cin>>a>>b>>c;
    for(int i=1;i<=b;i++)
    {
        cin>>x[i].s>>x[i].e>>x[i].ef;
        x[i].e+=c;
    }
    sort(x+1,x+1+b,cmp);
    memset(dp,0,sizeof(dp));
    int ans=0;
    for(int i=1;i<=b;i++)
    {
        dp[i]=x[i].ef;
        for(int j=1;j<i;j++)
        {
            if(x[i].s>=x[j].e)
            {
                dp[i]=max(dp[i],dp[j]+x[i].ef);//判断选这个或者不选这个，根据工作量大小
            }
        }
        ans=max(ans,dp[i]);
    }
    cout<<ans;
    
}