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题意：将9*9的棋盘分割成了9个部分，每个部分都是9个格子，然后现在要求每个部分的数字恰是1~9的排列，每一行每一列恰是1~9的排列，问是否有解，有多少组解，如果只有1组解打印出来
思路：先通过DFS求出所在的部分，然后剩下的和DLX精确覆盖求数独就是一样的了
#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck printf("fuck")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;const int MS = 10 + 5;
const int MX = 1000 + 5;
const int MN = 300000 + 5;
const int INF = 0x3f3f3f3f;int vis[3][MS][MS], Z[MS][MS], r;
int A[MS][MS], W[MS][MS], pre[MS][MS];struct DLX {int m, n;int H[MX], S[MX], ans;int Row[MN], Col[MN], rear;int L[MN], R[MN], U[MN], D[MN];void Init(int _m, int _n) {m = _m; n = _n;rear = n; ans = 0;for(int i = 0; i <= n; i++) {S[i] = 0;L[i] = i - 1;R[i] = i + 1;U[i] = D[i] = i;}L[0] = n; R[n] = 0;for(int i = 1; i <= m; i++) {H[i] = -1;}}void Link(int r, int c) {int rt = ++rear;Row[rt] = r; Col[rt] = c; S[c]++;D[rt] = D[c]; U[D[c]] = rt;U[rt] = c; D[c] = rt;if(H[r] == -1) {H[r] = L[rt] = R[rt] = rt;} else {int id = H[r];R[rt] = R[id]; L[R[id]] = rt;L[rt] = id; R[id] = rt;}}void Remove(int c) {R[L[c]] = R[c]; L[R[c]] = L[c];for(int i = D[c]; i != c; i = D[i]) {for(int j = R[i]; j != i; j = R[j]) {D[U[j]] = D[j]; U[D[j]] = U[j];S[Col[j]]--;}}}void Resume(int c) {for(int i = U[c]; i != c; i = U[i]) {for(int j = L[i]; j != i; j = L[j]) {D[U[j]] = U[D[j]] = j;S[Col[j]]++;}}R[L[c]] = L[R[c]] = c;}bool Dance(int cnt) {if(R[0] == 0) {for(int i = 1; i <= 9; i++) {for(int j = 1; j <= 9; j++) {if(pre[i][j]) W[i][j] = pre[i][j];}}ans++;if(ans >= 2) return true;return false;}int c = R[0];for(int i = R[0]; i != 0; i = R[i]) {if(S[i] < S[c]) c = i;}Remove(c);for(int i = D[c]; i != c; i = D[i]) {for(int j = R[i]; j != i; j = R[j]) Remove(Col[j]);int r = Row[i];pre[(r - 1) / 81 + 1][((r - 1) % 81) / 9 + 1] = ((r - 1) % 81) % 9 + 1;if(Dance(cnt + 1)) return true;for(int j = L[i]; j != i; j = L[j]) Resume(Col[j]);}Resume(c);return false;}
} G;void Link(int x, int y, int z, int k) {int id = ((x - 1) * 9 + y - 1) * 9 + k;G.Link(id, (x - 1) * 9 + y);G.Link(id, 9 * 9 + (x - 1) * 9 + k);G.Link(id, 2 * 9 * 9 + (y - 1) * 9 + k);G.Link(id, 3 * 9 * 9 + (z - 1) * 9 + k);
}void DFS(int x, int y, int id) {int w = A[x][y];Z[x][y] = id;if(w >= 128) w -= 128;else if(y - 1 >= 1 && !Z[x][y - 1]) DFS(x, y - 1, id);if(w >= 64) w -= 64;else if(x + 1 <= 9 && !Z[x + 1][y]) DFS(x + 1, y, id);if(w >= 32) w -= 32;else if(y + 1 <= 9 && !Z[x][y + 1]) DFS(x, y + 1, id);if(w >= 16) w -= 16;else if(x - 1 >= 1 && !Z[x - 1][y]) DFS(x - 1, y, id);W[x][y] = w;vis[0][x][w] = vis[1][y][w] = vis[2][id][w] = 1;
}int main() {int T, ansk = 0; //FIN;scanf("%d", &T);while(T--) {r = 0;memset(Z, 0, sizeof(Z));memset(pre, 0, sizeof(pre));memset(vis, 0, sizeof(vis));G.Init(9 * 9 * 9, 4 * 9 * 9);for(int i = 1; i <= 9; i++) {for(int j = 1; j <= 9; j++) {scanf("%d", &A[i][j]);}}for(int i = 1; i <= 9; i++) {for(int j = 1; j <= 9; j++) {if(!Z[i][j]) DFS(i, j, ++r);}}for(int i = 1; i <= 9; i++) {for(int j = 1; j <= 9; j++) {if(W[i][j]) Link(i, j, Z[i][j], W[i][j]);else for(int k = 1; k <= 9; k++) {int z = Z[i][j];if(!vis[0][i][k] && !vis[1][j][k] && !vis[2][z][k]) {Link(i, j, z, k);}}}}G.Dance(0);printf("Case %d:\n", ++ansk);if(G.ans == 0) printf("No solution\n");else if(G.ans == 2) printf("Multiple Solutions\n");else for(int i = 1; i <= 9; i++) {for(int j = 1; j <= 9; j++) {printf("%d", W[i][j]);}printf("\n");}}return 0;
}