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
題意：這個人要在n個小時內生產最多量的牛奶，兩個時間段不能重疊。典型的動態規劃，在處理到每一個時間段的時候，都要回過頭去遍歷一下前面的那些時間段。然后來更新。
代碼如下：

#include<iostream> #include<algorithm> #include<cstring> using namespace std; struct node{//定義一個結構體來儲存開始時間，結束時間，以及產生的價值int stime;int etime;int effect; }; struct node p[1010]; int dp[1010];//代表從開始到這個時間段可以生產的牛奶最大量 bool cmp(const node a,node b)//按著開始時間來排序 {return a.stime<b.stime; } int main() {int n,m,k;cin>>n>>m>>k;for(int i=0;i<m;i++){cin>>p[i].stime>>p[i].etime>>p[i].effect;}sort(p,p+m,cmp);//排序for(int i=0;i<m;i++){dp[i]=p[i].effect;//一開始定義為這樣，下面的循環開始更新for(int j=0;j<i;j++){if(p[j].etime+k>p[i].stime) continue;dp[i]=max(dp[i],dp[j]+p[i].effect);//更新}}int maxx=0;for(int i=0;i<m;i++){if(dp[i]>maxx)maxx=dp[i];}cout<<maxx<<endl;return 0; }

努力加油a啊，(^o)/~

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