2018 Multi-University Training Contest 4

dfs

每个4x4的宫格转了4次就变回原样了，所以最多转3次。

枚举没4行内的子矩阵，旋转后判断每行每列数是否冲突，在每次循环结束后子矩阵相当于没有转，因此遍历了每种情况。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
int _, ans;
char s[20][20];
bool vis[100];

int h(char ch){
    return isdigit(ch) ? ch - '0' : ch - 'A' + '0';
}

void rotate(int x, int y){
    char t[5][5];
    for(int i = 1, a = x; i <= 4; i ++, a ++){
        for(int j = 1, b = y; j <= 4; j ++, b ++) t[i][j] = s[a][b];
    }
    for(int j = 1, a = x; j <= 4; j ++, a ++){
        for(int i = 4, b = y; i >= 1; i --, b ++) s[a][b] = t[i][j];
    }
}

bool check(int row){
    for(int i = row; i <= row + 3; i ++){
        full(vis, false);
        for(int j = 1; j <= 16; j ++){
            if(vis[h(s[i][j])]) return false;
            vis[h(s[i][j])] = true;
        }
    }
    for(int j = 1; j <= 16; j ++){
        full(vis, false);
        for(int i = 1; i <= row + 3; i ++){
            if(vis[h(s[i][j])]) return false;
            vis[h(s[i][j])] = true;
        }
    }
    return true;
}

void dfs(int k, int step){
    if(step > ans) return;
    if(k > 16){
        ans = min(ans, step);
        return;
    }
    for(int a = 0; a <= 3; a ++, rotate(k, 1)){
        for(int b = 0; b <= 3; b ++, rotate(k, 5)){
            for(int c = 0; c <= 3; c ++, rotate(k, 9)){
                for(int d = 0; d <= 3; d ++, rotate(k, 13)){
                    if(!check(k)) continue;
                    dfs(k + 4, a + b + c + d + step);
                }
            }
        }
    }
}

int main(){
    for(_ = read(); _; _ --){
        for(int i = 1; i <= 16; i ++) scanf("%s", s[i] + 1);
        ans = INF;
        dfs(1, 0);
        printf("%d
", ans);
    }
    return 0;
}