1.题目是把一些数，通过组合，变成一个最小的数（在这些数的所有组合中最小）
2.使用贪心算法，贪心标准为：
1）如果两个数中，数a是数b的前缀，或者数b是数a的前缀，那么判断字符串相加后a+b与b+a的大小，进行确定前后顺序
2）如果不是上面1）中的两种情况，直接按照string进行大小比较
3.该题目可以通过微小的修改，改为求最大的数。
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
AC代码：
#include <iostream>
#include <stdio.h>
#include <vector>
#include <stack>
#include <algorithm>
#include <memory.h>
#include "limits.h"
using namespace std;
bool cmp(const string&a,const string&b)
{
//如果a和b，是前缀关系是，要比较a＋b与b+a的大小
if(a.size()<b.size()&&a==b.substr(0,a.size()))
return a+b<b+a;
else if(b.size()<a.size() && b==a.substr(0,b.size()))
return a+b<b+a;
else return a<b;
}
/*
5 32 321 3214 0229 87
4 11 122 12 10
4 122 12 10 11
*/
int main(void)
{
int n;
cin>>n;
vector<string> str(n);
for(int i=0;i<n;i++)
{
cin>>str[i];
}
sort(str.begin(), str.end(),cmp);
string ans="";
for(int i=0;i<n;i++)
{
ans+=str[i];
}
int i=0;
while(ans[i] == '0')
{
i++;
}
//注意全为0的状态
if(ans.substr(i)=="") cout<<0<<endl;
else cout<<ans.substr(i)<<endl;
return 0;
}