POJ 3616 Milking Time

Milking Time
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 5702 Accepted: 2386

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

先将时间段从小到大排好顺序，每次劳动完过后都要休息，每次都按顺序拿，每一次都和之前相比较，（现有的最优的情况dp[i]里已经存的值）和（a[i]自身这一个和之前的某一个相加之和）

for(int i=1;i<=m;i++)    {        dp[i]=a[i].number;        for(int j=1;j<i;j++)        {            if(a[i].start>=a[j].last+r)                dp[i]=max(dp[i],dp[j]+a[i].number);        }        ans=max(ans,dp[i]);    }

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;struct node{    int start;    int last;    int number;}a[1005];bool cmp(struct node p,struct node q){    if(p.start!=q.start)        return p.start<q.start;    return p.last<q.last;}int dp[1005];int main(void){   // freopen("D.txt","r",stdin);    int n,m,r;    scanf("%d%d%d",&n,&m,&r);    for(int i=1;i<=m;i++)        scanf("%d%d%d",&a[i].start,&a[i].last,&a[i].number);        sort(a+1,a+m+1,cmp);    dp[0]=0;    memset(dp,0,sizeof(dp));    int ans=-1;    for(int i=1;i<=m;i++)    {        dp[i]=a[i].number;        for(int j=1;j<i;j++)        {            if(a[i].start>=a[j].last+r)                dp[i]=max(dp[i],dp[j]+a[i].number);        }        ans=max(ans,dp[i]);    }      printf("%d\n",ans);    return 0;}