定义：dp[i] := 在已排好序的list的第 i 次John可从Bessie获取的最大牛奶量
dp[i] = max(dp[i], myNode[i].myvalue + dp[j]);
// 条件是： myNode[j].myend + R <= myNode[i].mystart （两重循环解决）
1 // [4/4/2014 Sjm]
2 #include <iostream>
3 #include <cstdio>
4 #include <algorithm>
5 using namespace std;
6 const int MAX_M = 1000, MAX_N = 1000001;
7 int N, M, R;
8 struct node {
9 int mystart, myend, myvalue;
10 };
11 bool Cmp(const node n1, const node n2) {
12 if (n1.mystart == n2.mystart) return n1.myend < n2.myend;
13 else return n1.mystart < n2.mystart;
14 }
15 node myNode[MAX_M];
16 int dp[MAX_M];
17
18 int Solve()
19 {
20 for (int i = 0; i < M; i++)
21 for (int j = 0; j < i; j++){
22 if (myNode[j].myend + R <= myNode[i].mystart){
23 dp[i] = max(dp[i], myNode[i].myvalue + dp[j]);
24 }
25 }
26 int ans = 0;
27 for (int i = 0; i < M; i++)
28 ans = max(ans, dp[i]);
29 return ans;
30 }
31
32 int main()
33 {
34 //freopen("input.txt", "r", stdin);
35 //freopen("output.txt", "w", stdout);
36 scanf("%d%d%d", &N, &M, &R);
37 for (int i = 0; i < M; i++) {
38 scanf("%d%d%d", &myNode[i].mystart, &myNode[i].myend, &myNode[i].myvalue);
39 }
40 sort(myNode, myNode + M, Cmp);
41 for (int i = 0; i < M; i++)
42 dp[i] = myNode[i].myvalue;
43 printf("%d
", Solve());
44 return 0;
45 }