2018_1_23_poj_1011_sticks_枚举_递归_剪枝

乔治将一些相同长度的棍子随机切割成不超过50单位的小段,现在他忘记了原始棍子的数量和长度。本篇介绍了一个算法帮助乔治找出这些棍子原本可能的最小长度。

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Sticks
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 150065 Accepted: 35693

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

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;

int sticks[65];
int used[65];
int n,len;

bool cmp(const int a,const int b){
	return a>b;
}

bool dfs(int i,int l,int t){
	if(l==0){
		t-=len;
		if(t==0)return true;
		for(i=0;used[i];i++);
		used[i]=1;
		if(dfs(i+1,len-sticks[i],t))return true;
		used[i]=0;t+=len;
	}else{
		for(int j=i;j<n;j++){
			if(j>0&&(sticks[j]==sticks[j-1]&&!used[j-1]))continue;
			if(!used[j]&&l>=sticks[j]){
				l-=sticks[j];used[j]=1;
				if(dfs(j,l,t))return true;
				l+=sticks[j];used[j]=0;
				if(sticks[j]==l)break;
			}
		}
	}
	return false;
}

int main(){
	while(cin>>n&&n){
		int sum=0;
		for(int i=0;i<n;i++){
			cin>>sticks[i];
			sum+=sticks[i];
			used[i]=0;
		}
		sort(sticks,sticks+n,cmp);
		bool flag=false;
		for(len=sticks[0];len<=sum/2;len++){
			if(sum%len==0){
				if(dfs(0,len,sum)){
					flag=true;
					cout<<len<<endl;
					break;
				}
			}
		}
		if(!flag)cout<<sum<<endl;
	}
	return 0;
}


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