Milking Time

 POJ - 3616 

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 Nhours

Sample Input

12 4 2
1 2 8
10 12 19
3 6 24
7 10 31

Sample Output

43

题目大意：第一行输入三个整数n,m,r，n代表最大时间,下面m行分别输入三个整数s,e,eff，表示从s时刻开始到e时刻结束共能收获eff的价值，注意每两段工作时间之间必须间隔r小时。

解决方法：先将其按照开始时间从小到大排序，然后求出dp[i]=max(dp[i],dp[j]+arr[i].eff)。

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <utility>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
struct node
{int s;int e;int eff;
}arr[1200];
bool cmp(node a,node b)
{if(a.s==b.s)return a.e<b.e;elsereturn a.s<b.s;
}
int dp[1200];
int main() 
{//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);ios::sync_with_stdio(0),cin.tie(0);int n,m,r;cin>>n>>m>>r;rep(i,1,m) {cin>>arr[i].s>>arr[i].e>>arr[i].eff;}sort(arr+1,arr+1+m,cmp);int ans=-1;for(int i=1;i<=m;i++){dp[i]=arr[i].eff;for(int j=1;j<i;j++){if(arr[j].e+r<=arr[i].s){dp[i]=max(dp[i],dp[j]+arr[i].eff);}}ans=max(ans,dp[i]);}cout<<ans<<endl;return 0;
}