poj1011 搜索

来源：互联网发布：网络存储软件有哪些编辑：程序博客网时间：2024/06/07 01:54

如题：http://poj.org/problem?id=1011

Sticks

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 125397 Accepted: 29215

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 should contains the smallest possible length of original sticks, one per line.

Sample Input

95 2 1 5 2 1 5 2 141 2 3 40

Sample Output

Source

Central Europe 1995

思路：搜索

1.注意可以搜索的拼成后的小木棍的长度的范围， max(a[i]) <=l<=sum(a[i])且sum%i==0

2.3个重要剪枝

1.某根木棒的长度不能构成，则之后的相同长度的也不行

2.如果在构成某根木棒的第一段时最大的无法用上，那么直接不可能

3.如果把某段拿去构成木棒的最后一截后，剩下的无法完成，那么也不可能

3.搜索时用vis数组记录这根小木棍是否已经使用，避免重复使用。

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

int n;
int a[70];
int vis[70];

int cmp(int x,int y)
{
return x>y;
}

int dfs(int total_len,int left_len,int left_num)
{
if(left_num==0&&left_len==0)
  return 1;
if(left_len==0)
  left_len=total_len;
int i;
for(i=0;i<n;i++)
{
  if(!vis[i])
  {
   if(left_len>=a[i])
   {
    vis[i]=1;
    if(dfs(total_len,left_len-a[i],left_num-1))
     return 1;
    vis[i]=0;
    if(left_len==total_len||left_len==a[i])
     break;
    while(a[i]==a[i+1])
     i++;
   }
  }
}
return 0;
}

int main()
{
// freopen("C:\\1.txt","r",stdin);
while(~scanf("%d",&n))
{
  if(!n)
   break;
  int sum=0;
  int i;
  for(i=0;i<n;i++)
  {
   scanf("%d",&a[i]);
   sum+=a[i];
  }
  sort(a,a+n,cmp);
  for(i=a[0];i<=sum;i++)
  {
   if(sum%i==0)
   {
   memset(vis,0,sizeof(vis));
   if(dfs(i,0,n))
    break;
   }
  }
  printf("%d\n",i);
}
}

0 0