题目描述
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.
奶牛Bessie在0~N时间段产奶。农夫约翰有M个时间段可以挤奶，时间段f,t内Bessie能挤到的牛奶量e。奶牛产奶后需要休息R小时才能继续下一次产奶，求Bessie最大的挤奶量。
输入输出格式
输入格式：
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
输出格式：
Line 1: The maximum number of gallons of milk that Bessie can product in the N hours
输入输出样例
输入样例#1：
12 4 2
1 2 8
10 12 19
3 6 24
7 10 31
输出样例#1：
43
题解
貌似这题是个区间dp吧，也貌似是个背包（排序之后）
思路：将区间按左端点从小到大的顺序排序，然后就变成了类似于求最长不下降子序列的问题。
假设dp[i]代表以i结尾的最大不重复区间和，那么动态转移方程就是：dp[i]=max(dp[j]+w[i])(1<=j< i)（其中w[i]表示区间i的价值,并且满足两区间不重叠，具体是：a[j].r+R<=a[i].l）
#include<bits/stdc++.h>
using namespace std;
struct wl{
int l,r,w;
}a[1001];
int n,m,R;
int dp[1001];
bool cmp(wl x,wl y){
return x.l<y.l;
}
int main(){
cin>>n>>m>>R;
for(int i=1;i<=m;i++){
scanf("%d%d%d",&a[i].l,&a[i].r,&a[i].w);
}
sort(a+1,a+m+1,cmp);
int ans=0;
for(int i=1;i<=m;i++){
dp[i]=a[i].w;
for(int j=1;j<i;j++){
if(a[j].r+R<=a[i].l){
dp[i]=max(dp[i],dp[j]+a[i].w);
}
}
ans=max(ans,dp[i]);
}
cout<<ans;
return 0;
}