POJ3616 Milking Time #最大上升子序列 基础DP#

版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。

本文链接： https://blog.csdn.net/qq_35850147/article/details/102485847

Milking Time Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17387   Accepted: 7467 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, 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 USACO 2007 November Silver

#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int maxm = 1e3 + 10;
int dp[maxm];
struct node { int beg, end, eff; } itv[maxm];

inline const int read()
{
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + ch - '0'; ch = getchar(); }
    return x * f;
}

bool cmp(const node& a, const node& b)
{
    return a.beg != b.beg ? a.beg < b.beg : a.end != b.end ? a.end < b.end : a.eff > b.eff;
}

int main()
{
    int n = read(), m = read(), r = read();
    for (int i = 0; i < m; i++)
    {
        itv[i].beg = read();
        itv[i].end = read();
        itv[i].eff = read();
    }
    sort(itv, itv + m, cmp);
    for (int i = 0; i < m; i++)
    {
        dp[i] = itv[i].eff;
        for (int j = 0; j < i; j++)
            if (itv[i].beg >= itv[j].end + r)
                dp[i] = max(dp[i], dp[j] + itv[i].eff);
    }
    int res = 0;
    for (int i = 0; i < m; i++) res = max(res, dp[i]);
    printf("%d\n", res);
    return 0;
}