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

题意：

　　给出一组数字，找出由这组数字组合形成的最小字符串。

思路：

　　先将这些数字按照字符串排序，然后依次输出，如果当前字符串和下一个字符串的子串相等，则比较下一个字符串的第一个字符和第len(current string)个字符的大小，输出较小的那个。然后将这个字符标记为已经访问过。下次循环再找出没有访问过的最小的字符串。

Code：

1 #include
2
3 using namespace std;
4
5 int main() {
6 int n;
7 cin >> n;
8 vector v(n + 1);
9 vector used(n, false);
10 for (int i = 0; i < n; ++i) cin >> v[i];
11 sort(v.begin(), v.begin() + n);
12 v.push_back("999999999");
13 string res;
14 int cnt = 0, i = 0, j;
15 while (cnt < n) {
16 j = i + 1;
17 while (used[j]) ++j;
18 int len = v[i].length();
19 if (v[i] == v[j].substr(0, len)) {
20 if (v[j][0] < v[j][len]) {
21 res += v[i];
22 used[i] = true;
23 while (used[i]) ++i;
24 } else {
25 res += v[j];
26 used[j] = true;
27 }
28 } else {
29 res += v[i];
30 used[i] = true;
31 while (used[i]) ++i;
32 }
33 cnt++;
34 }
35 int start = 0;
36 while (res[start] == '0') start++;
37 if (start == res.length()) cout << 0 << endl;
38 else cout << res.substr(start) << endl;
39 return 0;
40 }

注意：

　　如果结果是0的话需要进行特殊判断。不然的话会卡第三组数据。