Sticks(hdu1455,深搜)

最新推荐文章于 2020-03-30 16:14:16 发布

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最新推荐文章于 2020-03-30 16:14:16 发布

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分类专栏： DFS 文章标签： dfs 深度优先搜索深搜

本文链接：https://blog.youkuaiyun.com/weixin_41183791/article/details/81126881

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本文针对HDU OJ上的Sticks问题进行了解析，通过剪枝技巧优化递归搜索过程，找到能够还原切割前木棍最小长度的有效算法。

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传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1455

Sticks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12642 Accepted Submission(s): 3872

Problem Description

George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the original state, but he forgot how many sticks he had originally and how long they were originally. Please help him and design a program which computes the smallest possible original length of those sticks. All lengths expressed in units are integers greater than zero.

Input

The input contains blocks of 2 lines. The first line contains the number of sticks parts after cutting, there are at most 64 sticks. The second line contains the lengths of those parts separated by the space. The last line of the file contains zero.

Output

The output file contains the smallest possible length of original sticks, one per line.

Sample Input

9 5 2 1 5 2 1 5 2 1 4 1 2 3 4 0

Sample Output

6 5

这道题目需要许多剪枝,让我体会到剪枝的重要性

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;

int a[65],vis[65],n,flag,sum,maxd;

void dfs(int x,int d,int num)///x:第几段小木棍,d:当前组合长度,num:用了几段木棍
{
    if(flag) return ;
    if(num==n)///用完所以木棍
    {
        flag=1;
        return ;
    }
    for(int i=0;i<n;i++)
    {
        if(!vis[i]&&d+a[i]<=maxd)
        {
            vis[i]=1;
            if(d+a[i]==maxd)
                dfs(x+1,0,num+1);
            else dfs(x+1,d+a[i],num+1);
            vis[i]=0;
            if(flag) return ;
            if(d==0) return ;            ///最重要的 不写 直接TLE  当递归回来前面长棍子没用到的话 之后也就用不到了
            while(i<n&&a[i]==a[i+1]) i++;///当前长木棍不可以,呢么同样长就跳过
        }
    }
}

int main()
{
    while(scanf("%d",&n)&&n)
    {
        sum=0,flag=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        sort(a,a+n,greater<int>());
        for(int i=a[0];i<=sum&&!flag;i++)
        {
            if(sum%i==0)///可以整除才有必要分析
            {
                maxd=i;///合并后的最大长度
                dfs(0,0,0);
                if(flag)
                    printf("%d\n",i);
            }
        }
    }
    return 0;
}