Codeforces Round #501 (Div. 3) B. Obtaining the String

输入两个长为n的字符串s和t,问s经过多少次变换后(只能相邻交换,即i与i+1交换,且最后输出i的下标)可以变成t

当时唯一没想到的就是a。。。。c变成c。。。。a的话怎么移动,其实只需要从后往前就可以了233333333

菜到WA哭

#include <bits/stdc++.h>

using namespace std;
const int MAXN = 55;

int n;

char s[MAXN], t[MAXN];

vector<int> v;

int main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> s[i];
    for (int i = 1; i <= n; i++)
        cin >> t[i];
    for (int i = 1; i <= n; i++)
    {
        if (s[i] == t[i])
            continue;
        bool ch = 0;
        for (int j = i + 1; j <= n; j++)
        {
            if (s[j] == t[i])
            {
                for (int k = j - 1; k >= i; k--)
                {
                    v.push_back(k);
                    swap(s[k], s[k + 1]);
                }
                ch = 1;
            }
            if (ch == 1)
                break;
        }
        if (!ch)
        {
            cout << -1 << endl;
            exit(0);
        }
    }
    cout << v.size() << endl;
    int sz = v.size();
    for (int i = 0; i < sz; i++)
        cout << v[i] << " ";
    cout << endl;
    return 0;
}

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转载自www.cnblogs.com/Fy1999/p/9399126.html