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：

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n;
    cin >> n;
    vector<string> v(n + 1);
    vector<bool> used(n, false);
    for (int i = 0; i < n; ++i) cin >> v[i];
    sort(v.begin(), v.begin() + n);
    v.push_back("999999999");
    string res;
    int cnt = 0, i = 0, j;
    while (cnt < n) {
        j = i + 1;
        while (used[j]) ++j;
        int len = v[i].length();
        if (v[i] == v[j].substr(0, len)) {
            if (v[j][0] < v[j][len]) {
                res += v[i];
                used[i] = true;
                while (used[i]) ++i;
            } else {
                res += v[j];
                used[j] = true;
            }
        } else {
            res += v[i];
            used[i] = true;
            while (used[i]) ++i;
        }
        cnt++;
    }
    int start = 0;
    while (res[start] == '0') start++;
    if (start == res.length()) cout << 0 << endl;
    else cout << res.substr(start) << endl;
    return 0;
}

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。