PAT (Advanced Level) 1038. Recover the Smallest Number (30) 串联最小字符串，排序

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 (<=10000) 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. Do not output leading zeros.

Sample Input:

5 32 321 3214 0229 87

Sample Output:

22932132143287

字符串排序。将这些整数看成字符串，要把这些字符串串联起来得到最小的字符串。

由于高位权重较大，故前缀越小的字符串越应该放前面。

所有字符串串联起来后，在整串中，对于任意两个字符串s1和s2，有两种情况：***s1%%%s2###或***s2%%%s1###。

我们可以直接比较s1+s2和s2+s1。若s1+s2<s2+s1，即两个字符串串联时，s1放前面得到的字符串较小，说明他们放到整串中时，s1也应该放s2前面，即***s1%%%s2###小于***s2%%%s1###。

故我们可以直接对所有字符串从小到大排序。

输出时注意去掉第一个非全0字符串前面的0，若全为全0字符串则输出0。

/*2015.7.24cyq*/#include <iostream>#include <vector>#include <string>#include <sstream>#include <algorithm>using namespace std;int str2int(const string s1){stringstream ss;ss<<s1;int k;ss>>k;return k;}bool cmp(const string s1,const string s2){return s1+s2<s2+s1;}int main(){int N;cin>>N;vector<string> a;string s;for(int i=0;i<N;i++){cin>>s;a.push_back(s);}sort(a.begin(),a.end(),cmp);int k=0;while(k<N&&(str2int(a[k])==0))//去掉所有全0的片段 k++;if(k==N){cout<<0;return 0;}if(k<N){//第一个不是全0的片段，若前面有0需去掉int len=a[k].size();int i=0;while(a[k][i]=='0')i++;while(i<len){cout<<a[k][i];i++;}k++;}while(k<N){cout<<a[k];k++;}return 0;}