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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 }