POJ1316 Milking Time【dp】

Milking Time

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5896 Accepted: 2462

Description

Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her nextN (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 intervali 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 restR (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 theN 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 theN hours

Sample Input

12 4 21 2 810 12 193 6 247 10 31

Sample Output

43

题意给定一个时间间隔N在这个时间间隔内有M段段时间给出这M段时间每段的工作量和开始终止时间求N时间最大工作量每段工作后要休息R（将R算在每个时间段的结尾）

将这M段时间按开始时间从小到大排序dp[i]表示以i时间最大工作量；

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>using namespace std;struct Node{int start;int end;int efficitive;}A[1010];int dp[1010];bool cmp(Node a,Node b){return a.start<b.start;}int Max(int a,int b){return a>b?a:b;}int main(){int N,M,R,i,j;while(scanf("%d%d%d",&N,&M,&R)!=EOF){for(i=0;i<M;++i){scanf("%d%d%d",&A[i].start,&A[i].end,&A[i].efficitive);A[i].end+=R;}stable_sort(A,A+M,cmp);int count;count=dp[0]=A[0].efficitive;for(i=1;i<M;++i){dp[i]=A[i].efficitive;for(j=i-1;j>=0;--j){if(A[i].start-A[j].end>=0){dp[i]=Max(dp[i],dp[j]+A[i].efficitive);}}count=Max(count,dp[i]);}printf("%d\n",count);}return 0;}