Problem J. Let Sudoku Rotate
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PS:注意啊啊啊啊啊啊,Minato是逆时针旋转,那我们为了旋转回去,应该顺时针。没注意,debug花了好长时间。check的时候是判断当前的旋转是不是与先前的矛盾,比如将黑色那块旋转,就判断旋转后是不是与红色部分矛盾。
//#include<bits/stdc++.h> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<bitset> #include<vector> #include<queue> #include<map> #include<string> #include<stack> #define ll long long #define P pair<int, int> #define PP pair<int,pair<int, int>> #define pb push_back #define mp make_pair #define pp pop_back #define lson root << 1 #define INF (int)2e9 + 7 #define rson root << 1 | 1 #define LINF (unsigned long long int)1e18 #define mem(arry, in) memset(arry, in, sizeof(arry)) using namespace std; int T, res; int a[20][20], c[4][4], d[4][4], vis[20], g[20][20], use[20]; void Get(int st, int ed) { for (int i = st, q = ed; i < st + 4; ++i, ++q) { for (int j = ed, p = st + 3; j < ed + 4; ++j, --p) g[i][j] = a[p][q]; } for (int i = st; i < st + 4; ++i) { for (int j = ed; j < ed + 4; ++j) a[i][j] = g[i][j]; } } bool check(int st, int ed) { for(int i = st; i < st + 4; ++i) { mem(use, 0); for(int j = 0; j < ed + 4; ++j) { if(use[a[i][j]]) return false; use[a[i][j]] = 1; } } for(int i = ed; i < ed + 4; ++i) { mem(use, 0); for(int j = 0; j < st + 4; ++j) { if(use[a[j][i]]) return false; use[a[j][i]] = 1; } } return true; } void DFS(int deep, int cnt) { if (deep > 15) { res = min(res, cnt); return; } if(cnt >= res) return; int st = (deep / 4) * 4; int ed = (deep % 4) * 4; for (int i = 0; i < 4; ++i) { if(i) Get(st, ed); if(check(st, ed)) DFS(deep + 1, cnt + i); } Get(st, ed); } int main() { cin >> T; while (T--) { char ch; for (int i = 0; i < 16; ++i) { for (int j = 0; j < 16; ++j) { cin >> ch; a[i][j] = isdigit(ch) ? ch - ‘0‘ : ch - ‘A‘ + 10; } } res = 3000; DFS(0, 0); printf("%d ", res); } return 0; }
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