​POJ - 3628：Bookshelf 2（DFS）

原题链接
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 Hi (1 ≤ Hi ≤ 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: Hi

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

题目描述

已知书架高度为b，和n头牛的高度，求使用牛堆起来的高度与书架高度差值最小的值

思路：简单深搜，对存储牛高度的数组，逐个搜索，每次递归记录上次访问数组的下标（i+1），最后记录最小差值

代码如下

#include <stdio.h>
#include <iostream>
#include <algorithm>

using namespace std;

int n,b,cow[25],mins;

void dfs(int k, int sum)
{
    int i;
    if(sum == b)
    {
        mins = 0;
        return;
    }
    if(sum > b)
    {
        if(sum-b < mins)
            mins = sum-b;
        return;
    }
    for(i = k;i < n;i++)
        dfs(i+1, sum+cow[i]);	
	//一定是i+1，而不是k+1，因为函数回溯回来时不一定访问的第k+1数组元素
	//回溯回来因为已经试探过了k+1的元素，那么就需要试探i++之后的元素i+1
    return;
}

int main()
{
    int i;
    scanf("%d %d",&n,&b);
    for(i = 0;i < n;i++)
        scanf("%d",&cow[i]);
    mins = INT_MAX;
    dfs(0,0);
    printf("%d\n",mins);
    return 0;
}