Milking Time

Time Limit: 1000MS

 

Memory Limit: 65536K

Total Submissions: 10434

 

Accepted: 4378

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

题意：

在n时间内，有m个时间段，每段时间的值为v,每一段结束以后要休息r时间才能继续。

求n时间内v的最大值。

先按s从小到大排序，而后求出每一项前的最大值，则dp[i]=maxx+dp[j]。

AC代码：

//#include
#include
#include
#include
#include
using namespace std;

const int MAXN=;
int dp[MAXN];

struct node{
int s,e,v;
}a[MAXN];

int cmp(node x, node y){
return x.s<y.s;
}

int main(){
ios::sync_with_stdio(false);
int n,m,r;
while(cin>>n>>m>>r){
for(int i=;i>a[i].s>>a[i].e>>a[i].v;
}
sort(a,a+m,cmp);
int res=;
for(int i=;i<m;i++){
int maxx=;
for(int j=;j<i;j++){
if(a[j].e+r<=a[i].s&&maxx<dp[j])
maxx=dp[j];
}
dp[i]=maxx+a[i].v;
res=max(res,dp[i]);
}
cout<<res<<endl;
}
return ;
}