hdoj1455 poj1011 nyoj293 Sticks【DFS+剪枝】

Sticks

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 128649		Accepted: 30149

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

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int stick[70];
int vis[70];
int cont,nowcount;
int len;
int num;
bool cmp(int a,int b)
{
return a>b;
}
bool dfs(int k,int leftlen)
{
if(leftlen==0)
{
nowcount++;
if(nowcount==cont)return true;
int i=0;
while(vis[i])i++;
vis[i]=1;
if(dfs(i+1,len-stick[i]))return true;
vis[i]=0;nowcount--;
return false;//若当前最长的木棍不匹配则不能匹配
}
for(int i=k;i<num;++i)
{
if(!vis[i]&&stick[i]<=leftlen)
{
if(stick[i]==stick[i-1]&&!vis[i-1])continue;//前一个相同长度的没有匹配则以后相同的必定不会匹配
vis[i]=1;
if(dfs(i+1,leftlen-stick[i]))return true;
vis[i]=0;if(leftlen-stick[i]==0)return false;//当一根木棍已经匹配但不满足条件则不能完成匹配
}
}
return false;
}
int main()
{
int i,j;
while(scanf("%d",&num),num)
{
int sum=0;
for(i=0;i<num;++i)
{
scanf("%d",&stick[i]);
sum+=stick[i];vis[i]=0;
}
sort(stick,stick+num,cmp);
int flag=0;vis[0]=1;
for(len=stick[0];len<=sum/2;++len)//len最大为sum的1/2
{
if(sum%len)continue;//sum必定为len的倍数
cont=sum/len;nowcount=0;
if(dfs(1,len-stick[0]))
{
printf("%d\n",len);
flag=1;break;
}
}
if(flag==0)printf("%d\n",sum);
}
return 0;
}