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Sticks
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 127771 | Accepted: 29926 |
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
9 5 2 1 5 2 1 5 2 1 4 1 2 3 4 0
Sample Output
6 5
Source
题目链接：http://poj.org/problem?id=1011
题目大意：有n根木棍。用它们拼成一些等长的木棍，求拼出的木棍的最短长度。
解题思路：这题的时间限制特别严格。
DFS+剪枝，剪枝较多。首先由多到少枚举木棍数目num。即从n到1，要满足木棍总长度是num的倍数，且拼出的长度要不小于最长的木棍长度，否则无法拼，搜索到答案后退出循环，保证求出的木棍长最短。
剪枝：1.木棍由长到短排序。
2.訪问过的木棍或者加上当前木棍长度后超过了目标长度，则跳过本次循环。
3.若当前木棍和上一根木棍长度同样而且上一根木棍没用到，则跳过本次循环。
4.dfs中标记開始木棍下标。
代码例如以下：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[66],vis[66];
int n,num,m;
bool p;
int cmp(int a,int b)
{
return a>b;
}
void dfs(int st,int cur,int cnt)
{
if(p||cnt==num)
{
p=true;
return ;
}
for(int i=st;i<n;i++)
{
if(vis[i]||cur+a[i]>m) //訪问过的木棍或者加上当前木棍长度后超过了目标长度，则跳过本次循环
continue;
if(i-1&&!vis[i-1]&&a[i]==a[i-1]) //若当前木棍和上一根木棍长度同样而且上一根木棍没用到，则跳过本次循环。
continue;
if(a[i]+cur==m)
{
vis[i]=1;
dfs(0,0,cnt+1);
vis[i]=0;
return; //循环里后面的值都在dfs中求过了。这里直接返回上一层
}
if(a[i]+cur<m)
{
vis[i]=1;
dfs(i+1,a[i]+cur,cnt);
vis[i]=0;
if(cur==0) //cur为0时,直接返回上一层
return ;
}
}
}
int main()
{
while(scanf("%d",&n)!=EOF&&n)
{
int sum=0;
p=false;
memset(vis,0,sizeof(vis));
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
}
sort(a,a+n,cmp);
for(num=n;num>=1;num--)
{
if(sum%num||a[0]>sum/num)
continue;
m=sum/num;
dfs(0,0,0);
if(p)
break;
}
printf("%d\n",m);
}
return 0;
}
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