pat 甲级 1038. Recover the Smallest Number (30)

1038. Recover the Smallest Number (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

思路：可以先考虑其中任意两个字符串s1,s2，若s1+s2<s2+s1,那么s1必须要排在s2的左边才能使得最终结果尽可能的小，所以存在s1+s2>s2+s1的情况就对调s1,s2的位置,最终一定存在稳定状态，使得整个数字最小。这个模拟过程有点像冒泡排序
AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<set>
#include<queue>
#include<map>
using namespace std;
#define INF 0x3f3f3f
#define N_MAX 10000+5
typedef long long ll;
string s[N_MAX];
int n;
string process(string s) {
    int i = ;
    while (i < s.size() && s[i] == '')i++;
    string ss = s.substr(i, s.size() - i);
    if (!ss.size())ss = "";
    return ss;
}
bool cmp(const string &s1,const string &s2) {
    return s1+s2 < s2+s1;
}
int main() {
    scanf("%d",&n);
    for (int i = ; i < n; i++)cin >> s[i];
    sort(s, s + n, cmp);
    string S="";
    for (int i = ; i < n;i++) {
        S+=s[i];
    }
    S = process(S);
    cout << S<<endl;
    return ;
}