POJ3616——Milking Time

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Bessie要在N小时内通过合理安排被挤奶的时间来最大化产奶量。 Farmer John提供M个挤奶时间段，每个时间段有不同的效率。 Bessie在经过一个时间段的挤奶后需要休息R小时才能再次挤奶。任务是确定Bessie在这N小时内能产生的最大奶量。

Milking Time

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4964		Accepted: 2076

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

Sample Output

Source

USACO 2007 November Silver

dp，采用记忆化搜索，按题目意思建边就行

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1000010;
int dp[1010];
int head[1010];
int w[1010];
int tot;

struct kode
{
	int next;
	int to;
}edge[1010 * 1010];

struct node
{
	int l, r, eff;
}hour[1010];

void addedge(int from, int to)
{
	edge[tot].to = to;
	edge[tot].next = head[from];
	head[from] = tot++;
}

int dfs(int u)
{
	if (dp[u])
	{
		return dp[u];
	}
	for (int i = head[u]; ~i; i = edge[i].next)
	{
		int v = edge[i].to;
		dp[u] = max(dp[u], dfs(v));
	}
	dp[u] += w[u];
	return dp[u];
}

int main()
{
	int n, m, r;
	while (~scanf("%d%d%d", &n, &m, &r))
	{
		memset (head, -1, sizeof(head));
		memset (dp, 0, sizeof(dp));
		tot = 0;
		for (int i = 1; i <= m; ++i)
		{
			scanf("%d%d%d", &hour[i].l, &hour[i].r, &hour[i].eff);
			w[i] = hour[i].eff;
		}
		for (int i = 1; i <= m; ++i)
		{
			for (int j = i + 1; j <= m; ++j)
			{
				if (hour[i].r + r <= hour[j].l)
				{
					addedge(i, j);
					// printf("%d -- %d\n", i, j);
				}
				else if (hour[j].r + r <= hour[i].l)
				{
					addedge(j, i);
					// printf("%d -- %d\n", j, i);
				}
			}
		}
		int ans = 0;
		for (int i = 1; i <= m; ++i)
		{
			ans = max(ans, dfs(i));
		}
		printf("%d\n", ans);
	}
	return 0;
}