1038 Recover the Smallest Number (30分)(贪心)

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

题目分析：贪心算法 没想明白 贪心算法利用局部最优而达到整个问题的最优解 
所以在compare中写的比较函数是 a+b与b+a进行比较

 1 #define _CRT_SECURE_NO_WARNINGS
 2 #include <climits>
 3 #include<iostream>
 4 #include<vector>
 5 #include<queue>
 6 #include<map>
 7 #include<stack>
 8 #include<algorithm>
 9 #include<string>
10 #include<cmath>
11 using namespace std;
12 bool compare(const string& a, const string& b)
13 {
14     return a + b < b + a;
15 }
16 int main()
17 {
18     vector<string> S;
19     int N;
20     cin >> N;
21     string s;
22     for (int i = 0; i < N; i++)
23     {
24         cin >> s;
25         S.push_back(s);
26     }
27     s = "";
28     sort(S.begin(), S.end(), compare);
29     for (auto it : S)
30         s += it;
31     while (s.length() && s.at(0) == ‘0‘)
32         s.erase(s.begin());
33     if (!s.length())cout << 0;
34     else
35         cout << s;
36 }

View Code

1038 Recover the Smallest Number (30分)(贪心)

原文：https://www.cnblogs.com/57one/p/12017997.html