POJ3616Milking Time(DP)

Milking Time

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9610 Accepted: 3992

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 21 2 810 12 193 6 247 10 31

Sample Output

43

Source

USACO 2007 November Silver

题目大意:在n长度的时间中，有m个片段，有起始时间有终止时间以及奖励，要求在n时间段内获得最多的奖励，同时每次决定完成一个片段的时候，要有R分钟的休息时间

解题思路:这道题是我比较少接触的一个类似区间范围的dp，他要求尽可能的获得奖励的同时，不能在两个片段中出现交集，所以应该用dp,dp[i]指的是第i时间段(将m个时间段按照起始时间头最早时间排序),然后找比他大的起始时间中是否会出现冲突，若是不发生冲突，则求此时的max，之后遍历m即可

#include<iostream>    #include<cstdio>  #include<stdio.h>  #include<cstring>    #include<cstdio>    #include<climits>    #include<cmath>   #include<vector>  #include <bitset>  #include<algorithm>    #include <queue>  #include<map> #define inf 9999999; using namespace std;struct milk{int l,r,sum;}a[1005];bool cmp(milk x,milk y){if(x.l==y.l)return x.r<y.r;return x.l<y.l;}int i;int dp[1005];int n,m,r,j;int main(){cin>>n>>m>>r;for(i=1;i<=m;i++){cin>>a[i].l>>a[i].r>>a[i].sum;a[i].r+=r;}sort(a+1,a+1+m,cmp);memset(dp,0,sizeof(dp));for(i=m;i>=1;i--){dp[i]=a[i].sum;for(j=i+1;j<=m;j++){if(a[j].l>=a[i].r)dp[i]=max(dp[i],dp[j]+a[i].sum);}}int maxx=0;for(i=1;i<=m;i++){maxx=max(maxx,dp[i]);}cout<<maxx<<endl;}