POJ - 3616Milking Time挤牛奶（动态规划）

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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
题意：输入n,m,t,在n小时个区间，有m，但是每两个 挤奶区间必须相隔t，且每个区间挤奶量都不同，求最多可以挤出多少奶
分析：先sort排序，起始时间优先从小到达，其次是结束时间；注意这是动态规划，不是背包，最后一个数组值也不一定是最大的。所以从开始时间最小的开始循环，如果有比它结束时间早，且间隔大于等于t的，就可以在这两个时间段都挤奶 。

#include
#include
#include
using namespace std;
struct node
{
	int a,b,c;
}l[1010];
int cmp(node x, node y)
{
    if(x.a!=y.a) 
	return x.a<y.a;
    if(x.b!=y.b) 
	return x.b<y.b;
    return x.c<y.c;
}
int main()
{
	int n,m,t,i,j;
	int dp[1010];
	//scanf("%d%d%d",&n,&m,&t);
	 while(~scanf("%d%d%d",&n,&m,&t))
{
	for(i=1;i<=m;i++)
	scanf("%d%d%d",&l[i].a,&l[i].b,&l[i].c);
	sort(l+1,l+1+m,cmp);
	for(i=1;i<=m;i++)
	dp[i]=l[i].c;//dp代表这个时间段的挤奶量
	int ans=0;
	  for(i=1;i<=m;i++)
        {
            for(j=1;j<i;j++)
             {
                 if(l[i].a>=l[j].b+t) 
				 dp[i]=max(dp[i],dp[j]+l[i].c);//减小复杂度； 
             }
            ans=max(ans,dp[i]);
        }
	printf("%d\n",ans);
	//printf("%d %d %d %d %d %d\n",dp[0],dp[1],dp[2],dp[3],dp[4],dp[5]);
 }
	
}