Poj 3616 Milking Time【dp】

Milking Time

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7144 Accepted: 2988

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

题意：

有个人的奶牛要产奶，给出这个人在某些时间段的能够采集的量，且某一时间段内不允许中断，问这个人最多能采集多少牛奶

动态规划不好做，一直找递推式，就是不知道到底如何下手，后来看了大神的题解，也是理解的迷迷糊糊的，真的是不会做.......

本题的dp[i]表示第 i 个时间段时一共能采集的量，对每个时间段都只统计它后边的时间段的最值，这样避免了很多的时间段的交叉的复杂的处理。

/*http://blog.csdn.net/liuke19950717*/#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{    int bg,ed,milk;}x[1005];int cmp(node a,node b){    if(a.ed==b.ed)    {        return a.bg<b.bg;    }    return a.ed<b.ed;}int slove(int n,int m,int r){    int dp[1005]={0};    sort(x,x+m,cmp);    for(int i=0;i<m;++i)    {        dp[i]=x[i].milk;}//dp[i] 表示第 i 次工作时，挤奶的最大量      for(int i=0;i<m;++i)//    {        for(int j=i+1;j<m;++j)        {            if(x[i].ed+r<=x[j].bg)            {            dp[j]=max(dp[j],dp[i]+x[j].milk);}        }    }    int ans=0;    for(int i=0;i<m;++i)    {    ans=max(ans,dp[i]);}    return ans;}int main(){    int n,m,r;    while(~scanf("%d%d%d",&n,&m,&r))    {        for(int i=0;i<m;++i)        {            scanf("%d%d%d",&x[i].bg,&x[i].ed,&x[i].milk);        }        printf("%d\n",slove(n,m,r));    }    return 0;}