RC-u4 变牛的最快方法

#include <bits/stdc++.h>
using namespace std;

const int N = 1010;

int a[N], b[N];
int f[N][N]; // a数组中前i个数字转变为b数组中前j个数字的最少变换次数
int ans[N][N]; //抵达该状态的最后一个操作类别
pair<int, int> pre[N][N]; //抵达该状态的前一个状态
int n, m;

int main()
{
    int x;
    while(cin >> x && x != -1) a[++ n] = x;
    while(cin >> x && x != -1) b[++ m] = x;

    //初始化
    f[0][0] = 0;

    //状态转移以及记录状态
    for(int i = 1; i <= n; i ++ ) f[i][0] = f[i - 1][0] + 1, ans[i][0] = 0, pre[i][0] = {-1, 0};
    for(int i = 1; i <= m; i ++ ) f[0][i] = f[0][i - 1] + 1, ans[0][i] = 3, pre[0][i] = {0, -1};

    //状态转移以及记录状态
    for(int i = 1; i <= n; i ++ )
        for(int j = 1; j <= m; j ++ )
        {
            if(a[i] == b[j]) f[i][j] = f[i - 1][j - 1], ans[i][j] = 2, pre[i][j] = {-1, -1};
            else
            {
                if(f[i - 1][j - 1] <= min (f[i - 1][j], f[i][j - 1]))
                {
                    f[i][j] = f[i - 1][j - 1] + 1;
                    ans[i][j] = 1;
                    pre[i][j] = {-1, -1};
                }
                else if(f[i - 1][j] <= min (f[i - 1][j - 1], f[i][j - 1]))
                {
                    f[i][j] = f[i - 1][j] + 1;
                    ans[i][j] = 0;
                    pre[i][j] = {-1, 0};
                }
                else
                {
                    f[i][j] = f[i][j - 1] + 1;
                    ans[i][j] = 3;
                    pre[i][j] = {0, -1};
                }
            }
        }

    //输出状态
    cout << f[n][m] << endl;
    vector<int> v;
    /*
    -1
    13 14 15 16 -1
    */
    while(n || m)
    {
        v.push_back(ans[n][m]);
        auto t = pre[n][m];
        n += t.first;
        m += t.second;
    }
    for(int i = v.size() - 1; i >= 0; i -- )
        cout << v[i];
    return 0;
}