PAT甲级——A1038 Recover the Smallest Number

Given a collection of number segments, you are supposed to recover the smallest number from them. For example, given { 32, 321, 3214, 0229, 87 }, we can recover many numbers such like 32-321-3214-0229-87 or 0229-32-87-321-3214 with respect to different orders of combinations of these segments, and the smallest number is 0229-321-3214-32-87.

Input Specification:

Each input file contains one test case. Each case gives a positive integer N (≤) followed by N number segments. Each segment contains a non-negative integer of no more than 8 digits. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the smallest number in one line. Notice that the first digit must not be zero.

Sample Input:

5 32 321 3214 0229 87

Sample Output:

22932132143287


 1 #include <iostream>
 2 #include <string>
 3 #include <vector>
 4 #include <algorithm>
 5 using namespace std;
 6 //排序问题
 7 int N;
 8 vector<string>v, temp;
 9 string res = "", str = "";
10 void permuteDFS(int u,vector<bool>&visit)//使用DFS
11 {
12     if (u == v.size())
13     {
14         for (auto a : temp)
15             str += a;
16         res = res > str ? str : res;
17         str = "";
18         return;
19     }
20     for (int i = 0; i < N; ++i)
21     {
22         if (visit[i] == true)continue;
23         visit[i] = true;
24         temp.push_back(v[i]);
25         permuteDFS(i + 1, visit);
26         temp.pop_back();
27         visit[i] = false;
28     }
29 }
30 
31 void permutex(int u)//使用递归
32 {
33     if (u == v.size())
34     {
35         for (auto a : v)
36             str += a;
37         res = res > str ? str : res;
38         str = "";
39     }
40     for (int i = u; i < N; ++i)
41     {
42         swap(v[i], v[u]);
43         permutex(i + 1);
44         swap(v[i], v[u]);
45     }
46 }
47 
48 void Sort()//使用排序法则
49 {
50     sort(v.begin(), v.end(), [](string a, string b) {return a + b < b + a; });
51     res = "";
52     for (auto a : v)
53         res += a;
54 }
55 
56 int main()
57 {
58     cin >> N;
59     v.resize(N);
60     vector<bool>visit(N, false);
61     for (int i = 0; i < N; ++i)
62     {
63         cin >> v[i];
64         res += v[i];
65     }
66     //permutex(0);
67     //permuteDFS(0, visit);
68     Sort();
69     while (!res.empty() && res[0] == '0')
70         res.erase(0, 1);
71     if (res.size() == 0)cout << 0;
72     cout << res << endl;
73     return 0;
74 }

 

posted @ 2019-07-27 22:25  自由之翼Az  阅读(172)  评论(0)  编辑  收藏  举报