Sticks（剪枝+BFS）

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.

SampleInput

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

SampleOutput

6
5

題意就是給你一堆不一樣長度的木棍，要你用他們拼成長度相等的木棍，輸出最短且知足要求的木棍長度。
就直接從0開始搜嘛，判斷長度相同就輸出，好了就這樣，提交，TLE。
=7=那麼看來不能直接搜索，如何優化呢，固然就是判斷一些條件剪枝，首先是搜索範圍，根據題意有原始長度確定比如今最長的要長，並且能被如今總長除盡，原始長度最長確定就是總長度（只有一根木棍），其次就是對於同一個長度的木棍，只嘗試一次，一次不成功便不須要嘗試第二次，而後就依次dfs就好了。
代碼：

 1 #include <iostream>
 2 #include <string>
 3 #include <cstdio>
 4 #include <cstdlib>
 5 #include <sstream>
 6 #include <iomanip>
 7 #include <map>
 8 #include <stack>
 9 #include <deque>
10 #include <queue>
11 #include <vector>
12 #include <set>
13 #include <list>
14 #include <cstring>
15 #include <cctype>
16 #include <algorithm>
17 #include <iterator>
18 #include <cmath>
19 #include <bitset>
20 #include <ctime>
21 #include <fstream>
22 #include <limits.h>
23 #include <numeric>
24 
25 using namespace std;
26 
27 #define F first
28 #define S second
29 #define mian main
30 #define ture true
31 
32 #define MAXN 1000000+5
33 #define MOD 1000000007
34 #define PI (acos(-1.0))
35 #define EPS 1e-6
36 #define MMT(s) memset(s, 0, sizeof s)
37 typedef unsigned long long ull;
38 typedef long long ll;
39 typedef double db;
40 typedef long double ldb;
41 typedef stringstream sstm;
42 const int INF = 0x3f3f3f3f;
43 
44 int s[500050];
45 bool vis[500050];
46 int sum,n;
47 
48 bool dfs(int pos, int res, int len){
49     if(pos == 0 && res == 0)
50         return true;
51     if(!res)
52         res = len;
53     for(int i = n - 1; i >= 0; i--){
54         if(vis[i] || s[i] > res)
55             continue;
56         vis[i] = true;
57         if(dfs(pos - 1, res - s[i], len))
58             return true;
59         vis[i] = false;
60         if(res == s[i] || res == len)    //判斷是否嘗試過
61             return false;
62     }
63     return false;
64 }
65 
66 int main(){
67     ios_base::sync_with_stdio(false);
68     cout.tie(0);
69     cin.tie(0);
70     while(cin >> n && n){
71         sum = 0;
72         for(int i = 0; i < n; i++){
73             cin >> s[i];
74             sum += s[i];
75         }
76         sort(s, s+n);
77         int i;
78         for(i = s[n - 1]; i <= sum/2;i++){    //從最大長度開始dfs
79             if(sum % i)　　//小於總長度則dfs嘗試
80                 continue;
81             MMT(vis);　　//每次dfs前記得清空標記
82             if(dfs(n,i,i)){
83                 cout << i << endl;
84                 break;
85             }
86         }
87         if(i > sum/2)    //大於長度一半，則不可能被分紅多個木棍，確定只能由一根木棍構成
88             cout << sum << endl;
89     }
90     return 0;
91 }