Description

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.

The input file may contain multiple test cases. The first line of each test case contains an integerN (1 ￡ N ￡ 1000) that denotes the number of books to shelve, and a floating-point numberW (0 < W ￡ 1000) that denotes the width of the bookcase in centimeters. Then followN lines where the i-th (1 ￡ i ￡ N) line contains two floating-point numbersh_i (0 < h_i ￡ 100) and w_i (0 < w_i ￡ W) indicating the height and width (both in centimeters) of thei-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.

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.

65.0000

题意：n本书，有宽和高，要你依次放到书架上，能够用木板隔层，每层最多不超过m的宽度

思路：居然要求是依次放，那么每本书有放这层和不放这层的两种可能，记忆化搜索，注意要考虑全都放一层的情况

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstdlib>
using namespace std;
const int maxn = 1005;
const double inf = 0x3f3f3f3f3f3f3f3f;
double m, h[maxn], w[maxn], dp[maxn];
int n, vis[maxn];

double dfs(int cur) {
	if (cur >= n+1)
		return 0;
	if (vis[cur])
		return dp[cur];
	double &ans = dp[cur];
	ans = inf;
	vis[cur] = 1;
	double H = h[cur], W = w[cur];
	for (int u = cur+1; u <= n+1; u++) {
		ans = min(ans, dfs(u)+H);
		W += w[u];
		H = max(H, h[u]);
		if (W-m > 1e-9)
			break;
	}
	return ans;
}

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