程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> POJ3616——Milking Time

POJ3616——Milking Time

編輯：C++入門知識

POJ3616——Milking Time

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，采用記憶化搜索，按題目意思建邊就行

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

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;
}