1038 Recover the Smallest Number

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 <bits/stdc++.h>
 2
 3 using namespace std;
 4
 5 int main() {
 6     int n;
 7     cin >> n;
 8     vector<string> v(n + 1);
 9     vector<bool> 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。