POJ 3616 Milking Time 擠奶問題，帶權區間DP
Milking Time
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 * 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 |
題意：
每一個擠奶區間都有一個效率，奶牛每擠完依次奶需要休息R小時，問怎樣安排才能使效率最高。
分析：
如果沒有效率問題，那麼按照結束時間遞增排序總是最優的，這樣可以覆蓋到儘可能多的區間，當加上效率時，我們要考慮到前面的時間獲得的效率是否比當前時間段的效率要更高。若以dp[k]表示第k個區間的效率（注意已經排好序，因而就是第k個結束時間的效率。則狀態轉移方程為：
dp[k] = max(dp[k], dp[j]+itv[j].ef);
程式碼：
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAX_N = 1000010;
const int MAX_M = 1010;
struct ITV {
int st, ed, ef;
bool operator < (const ITV& b) const {
if(ed == b.ed) return st < b.st;
return ed < b.ed;
}
}itv[MAX_M];
int N, M, R, dp[MAX_M];
void solve() {
sort(itv, itv+M);
for(int i = 0; i < M; i++) dp[i] = itv[i].ef;
int res = 0;
for(int i = 0; i < M; i++)
for(int j = 0; j < i; j++)
if(itv[i].st - itv[j].ed >= R) {
dp[i] = max(dp[i], dp[j]+itv[i].ef);
if(dp[i] > res) res = dp[i];
}
printf("%d\n", res);
}
int main() {
while(~scanf("%d%d%d", &N, &M, &R)) {
for(int i = 0; i < M; i++)
scanf("%d%d%d", &itv[i].st, &itv[i].ed, &itv[i].ef);
solve();
}
return 0;
}