Bookshelf 2

Time Limit:?1000MS

?

Memory Limit:?65536K

Description

Farmer John recently bought another bookshelf for the cow library, but the shelf is getting filled up quite quickly, and now the only available space is at the top.

FJ has?N?cows (1 ≤?N?≤ 20) each with some height of?H_i?(1 ≤?H_i?≤ 1,000,000 - these are very tall cows). The bookshelf has a height of?B?(1 ≤?B?≤?S, where?S?is the sum of the heights of all cows).

To reach the top of the bookshelf, one or more of the cows can stand on top of each other in a stack, so that their total height is the sum of each of their individual heights. This total height must be no less than the height of the bookshelf in order for the cows to reach the top.

Since a taller stack of cows than necessary can be dangerous, your job is to find the set of cows that produces a stack of the smallest height possible such that the stack can reach the bookshelf. Your program should print the minimal 'excess' height between the optimal stack of cows and the bookshelf.

Input

* Line 1: Two space-separated integers:?N?and?B
* Lines 2..N+1: Line?i+1 contains a single integer:?H_i

Output

* Line 1: A single integer representing the (non-negative) difference between the total height of the optimal set of cows and the height of the shelf.

Sample Input

5 16 3 1 3 5 6

Sample Output

1

題意：有n頭牛，已知每頭牛的高度和書架高度，一頭牛可以站在另一頭牛身上，總高度是他們的高度之和。要求能夠達到書架的頂端，即這些牛的總高度不低于書架高度，求滿足條件的總高度的最小值，輸出他們的差值。

解題思路：把書架的高度看做背包容量，將牛的高度看做物品的體積和價值，dp[i] 表示書架高度為i，牛的總高度不超過i的最大高度。

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int dp[2000002], h[22]; int main() {int n, m, i, j;while(~scanf("%d%d",&n,&m)){int sum = 0;memset(dp,0,sizeof(dp));for(i = 1; i <= n; i++){scanf("%d",&h[i]);sum += h[i];}for(i = 1; i <= n; i++)for(j = sum; j >= h[i]; j--)dp[j] = max(dp[j], dp[j - h[i]] + h[i]);int Min = sum;for(i = m; i <= sum; i++)if(dp[i] >= m && dp[i] - m < Min)Min = dp[i] - m;printf("%d\n",Min);}return 0; }
也可以用搜索來寫：

#include<cstdio> #include<cstring> #include<iostream> using namespace std; int h[22], ans, flag; int n, m; void dfs(int k, int s) {if(s == m){ans = 0;return ;}if(s >= m){if(s - m < ans)ans = s - m;return ;}for(int i = k; i < n; i++){dfs(i+1,s+h[i]);} } int main() {int i;while(cin >> n >> m){int sum = 0;flag = 0;for(i = 0; i < n; i++){cin >> h[i];sum += h[i];}if(sum == m){cout << "0" << endl;continue;}ans = sum;dfs(0,0);cout << ans << endl;}return 0; }

總結

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