#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;
}
RC-u4 变牛的最快方法
于 2024-07-31 13:47:07 首次发布