poj 3616 Milking Time（dp）

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

 

 

 

dp，注意初始化，

for(int i=1;i<=m;i++){
dp[i]=cows[i].c;
}

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stdlib.h>
using namespace std;
#define N 1000006
#define M 1006
int n,m,r;
struct Node{
    int s,e;
    int c;
}cows[M];
bool cmp(Node a,Node b){
    if(a.s!=b.s)
      return a.s<b.s;
     return a.e<b.e;
}
int dp[M];//dp[i]表示取到第i段时间段时的最大值 
int main()
{
    while(scanf("%d%d%d",&n,&m,&r)==3){
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&cows[i].s,&cows[i].e,&cows[i].c);
        }
        sort(cows+1,cows+m+1,cmp);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=m;i++){
            dp[i]=cows[i].c;
        }
        //dp[1]=cows[1].c;
        int ans=0;
        for(int i=1;i<=m;i++){
            for(int j=1;j<i;j++){
                if(cows[i].s>=cows[j].e+r){
                    dp[i]=max(dp[i],dp[j]+cows[i].c);
                }
            }
            ans=max(dp[i],ans);
        //    printf("---%d\n",ans);
        }
        printf("%d\n",ans);
    }
    return 0;
}