Milking Time
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
Source
题目类型：线性DP
算法分析：dp[i]表示以第i个时间段为结尾所能够挤出的最大奶量，类似最长递增子序列的递推形式
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/************************************************* Author :supermaker Created Time :2015/12/2 14:38:24 File Location :C:\Users\abcd\Desktop\TheEternalPoet **************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e6 + 666; LL dp[maxn]; struct node { LL a, b, c; }; bool operator < (const node&va, const node&vb) { if (va.a != vb.a) return va.a < vb.a; return va.b < vb.b; } node aa[maxn]; int main() { //CFF; //CPPFF; int N, M, R; while (scanf ("%d%d%d", &N, &M, &R) != EOF) { for (int i = 1; i <= M; i++) scanf("%lld%lld%lld", &aa[i].a, &aa[i].b, &aa[i].c); sort (aa + 1, aa + 1 + M); dp[1] = aa[1].c; for (int i = 2; i <= M; i++) { LL maxvala = 0; for (int j = 1; j <= i - 1; j++) if (aa[j].b + R <= aa[i].a) maxvala = max (maxvala, dp[j]); dp[i] = maxvala + aa[i].c; } printf ("%lld\n", *max_element (dp + 1, dp + 1 + M)); } return 0; } |
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