UVA 607 (dp记忆化搜索)

最新推荐文章于 2020-10-13 22:42:41 发布

原创最新推荐文章于 2020-10-13 22:42:41 发布 · 1.1k 阅读

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本文探讨了在教学过程中如何高效安排课程与讲座，确保覆盖所有必要主题的同时，通过减少讲座间的空闲时间来提高学生满意度。文章提出了基于贪心算法与记忆化搜索的方法，以最小化讲座数量并降低总不满意度指数。

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Scheduling Lectures

You are teaching a course and must cover n ( $1 \le n \le 1000$ ) topics. The length of each lecture is L ( $1 \le L \le 500$ ) minutes. The topics require $t_1, t_2, \dots, t_n$ ( $1 \le t_i \le L$ ) minutes each. For each topic, you must decide in which lecture it should be covered. There are two scheduling restrictions:

Each topic must be covered in a single lecture. It cannot be divided into two lectures. This reduces discontinuity between lectures.

Topic i must be covered before topic i + 1 for all $1 \le i < n$ . Otherwise, students may not have the prerequisites to understand topic i + 1.

With the above restrictions, it is sometimes necessary to have free time at the end of a lecture. If the amount of free time is at most 10 minutes, the students will be happy to leave early. However, if the amount of free time is more, they would feel that their tuition fees are wasted. Therefore, we will model the dissatisfaction index (DI) of a lecture by the formula:

$\begin{displaymath}DI = \cases{0 & if $t=0$, \cr-C & if $1 \le t \le 10$, \cr(t-10)^2 & otherwise}\end{displaymath}$

where C is a positive integer, and t is the amount of free time at the end of a lecture. The total dissatisfaction index is the sum of the DI for each lecture.

For this problem, you must find the minimum number of lectures that is needed to satisfy the above constraints. If there are multiple lecture schedules with the minimum number of lectures, also minimize the total dissatisfaction index.

Sample Output

Case 1:
Minimum number of lectures: 2
Total dissatisfaction index: 0

Case 2:
Minimum number of lectures: 6
Total dissatisfaction index: 2700

题意：给n节课时间，然后有一个一课时时间ti，和一个常数c，课必须连续着上不能换顺序，要求出最少需要的课时，和在该课时下学生不满意度最小值。

思路：前一步利用贪心，可以算出最少课时，就是尽量让多的课挤到一起上，然后在用记忆化搜索，求出该课时下最小满意度

代码：

#include <stdio.h>
#include <string.h>
const int N = 1005;
const int INF = 1 << 30;
int min(int a, int b) {return a < b ? a : b;}
int n, ti, c, count, t[N], dp[N][N], vis[N][N], i, j, num; 

int get(int x) {
	if (x == 0) return 0;
	else if (x <= 10) return -c;
	else return (x - 10) * (x - 10);
}

int solve(int x, int y) {
	if (vis[x][y]) return dp[x][y];
	int &ans = dp[x][y];
	if (x == 0) return y ? INF : 0;
	else {
		ans = INF;
		int sum = 0;
		for (int i = y; i > 0; i --) {
			sum += t[i];
			if (sum > ti) break;
			int now = solve(x - 1, i - 1);
			if (now != INF)
				ans = min(now + get(ti - sum), ans);
		}
		vis[x][y] = 1;
		return ans;
	}
}

int main() {
	int tt = 0;
	while (~scanf("%d", &n) && n) {
		num = 0;
		int sum = 1;
		memset(vis, 0, sizeof(vis));
		scanf("%d%d", &ti, &c);
		for (i = 1; i <= n; i ++) {
			scanf("%d", &t[i]);
			num += t[i];
			if (num > ti) {
				num = t[i];
				sum ++;
			}
		}
		if (tt) printf("\n");
		printf("Case %d:\n", ++ tt);
		printf("Minimum number of lectures: %d\n", sum);
		printf("Total dissatisfaction index: %d\n", solve(sum, n));
	}
	return 0;
}