UVA 10239 - The Book-shelver's Problem(记忆化搜索)

Problem D
The Book-shelver’s Problem
Input: standard input
Output: standard output
Time Limit: 5 seconds
Memory Limit: 32 MB

 

You are given a collection of books, which must be shelved in a library bookcase ordered (from top to bottom in the bookcase and from left to right in each shelf) by the books’ catalogue numbers. The bookcase has a fixed width, but you may have any height you like. The books are placed on shelves in the bookcase in the usual upright manner (i.e., you cannot lay a book on its side). You may use as many shelves as you like, placed wherever you like up to the height of the bookcase, and you may put as many books on each shelf as you like up to the width of the bookcase. You may assume that the shelves have negligible thickness.

 

Now, given an ordered (by catalogue numbers) list of the heights and widths of the books and the width of the bookcase, you are expected to determine what is the minimum height bookcase that can shelve all those books.

 

Input

The input file may contain multiple test cases. The first line of each test case contains an integer N (1 £ N £ 1000) that denotes the number of books to shelve, and a floating-point number W (0 < W £ 1000) that denotes the width of the bookcase in centimeters. Then follow N lines where the i-th (1 £ i £ N) line contains two floating-point numbers h_i (0 < h_i £ 100) and w_i (0 < w_i £ W) indicating the height and width (both in centimeters) of the i-th book in the list ordered by catalogue numbers. Each floating-point number will have four digits after the decimal point.

 

A test case containing two zeros for N and W terminates the input.

 

Output

For each test case in the input print a line containing the minimum height (in centimeters, up to four digits after the decimal point) of the bookcase that can shelve all the books in the list.

 

Sample Input
5 30.0000
30.0000 20.0000
20.0000 10.0000
25.0000 10.0000
30.0000 15.0000
10.0000 5.0000
10 20.0000
10.0000 2.0000
15.0000 10.0000
20.0000 5.0000
6.0000 2.0000
10.0000 3.0000
30.0000 6.0000
5.0000 3.0000
35.0000 2.0000
32.0000 4.0000
10.0000 6.0000
0 0.0000

 

Sample Output
60.0000
65.0000

题意：一个书架宽度为W，现在n本书，书有宽度高度，现在按顺序一本本去放书，现在有很多架子，在放书的时候可以选择利用架子换一层去放，或者放在当前一层，但是宽度不能超过W。问放完书的最低高度

思路：记忆化搜索，dp[i]表示放i本书之后的书需要的最小高度，那么一本本去放书，书要嘛换层要嘛继续放（只要小于W）。

代码：

#include <stdio.h>
#include <string.h>
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
const int N = 1005;
int n, vis[N];
double W, h[N], w[N], dp[N];

double dfs(int start) {
	if (vis[start]) return dp[start];
	vis[start] = 1;
	dp[start] = 1000000000;
	if (start > n) return dp[start] = 0;
	double ws = 0, hmax = 0;
	for (int i = start; i <= n; i++) {
		ws += w[i];
		hmax = max(hmax, h[i]);
		if (ws - W > 1e-6) break;
		double v = dfs(i + 1);
		dp[start] = min(dp[start], v + hmax);
	}
	return dp[start];
}

int main() {
	while (~scanf("%d%lf", &n, &W) && n) {
		memset(vis, 0, sizeof(vis));
		for (int i = 1; i <= n; i++)
			scanf("%lf%lf", &h[i], &w[i]);
		printf("%.4lf\n", dfs(1));
	}
	return 0;
}