POJ3616 Milking Time 【DP】

Milking Time

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4917 Accepted: 2062

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_hour_i ≤ N), an ending hour (starting_hour_i <ending_hour_i ≤ N), and a corresponding efficiency (1 ≤ efficiency_i ≤ 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_hour_i , ending_hour_i , and efficiency_i

Output

* Line 1: The maximum number of gallons of milk that Bessie can product in the N hours

Sample Input

12 4 21 2 810 12 193 6 247 10 31

Sample Output

43

Source

USACO 2007 November Silver

最基础的DP，将区间按照左端排序，容易想到状态是dp[i]表示以第i个区间作为结束区间能得到的最大值。

#include <stdio.h>#include <string.h>#include <algorithm>#define maxn 1010using namespace std;struct Node {    int u, v, w;    friend bool operator<(const Node& a, const Node& b) {        return a.u < b.u;    }} E[maxn];int N, M, R, dp[maxn]; // dp[i]表示以第i个区间作为结束区间能得到的最大值int main() {    int ans, i, j;    while(scanf("%d%d%d", &N, &M, &R) == 3) {        for(i = 0; i < M; ++i)            scanf("%d%d%d", &E[i].u, &E[i].v, &E[i].w);        sort(E, E + M);        ans = dp[0] = E[0].w;        for(i = 1; i < M; ++i) {            dp[i] = E[i].w;            for(j = i - 1; j >= 0; --j)                if(E[i].u - E[j].v >= R)                    dp[i] = max(dp[i], dp[j] + E[i].w);            ans = max(ans, dp[i]);        }        printf("%d\n", ans);    }    return 0;}