題目鏈接
題意:中文題
思路:每一個1看成列,每個位置作為左上角的矩陣看成行,dlx重復覆蓋即可
代碼:
#include#include using namespace std; const int MAXNODE = 66666; const int INF = 0x3f3f3f3f; const int MAXM = 230; const int MAXN = 230; int K; struct DLX { int n,m,size; int U[MAXNODE], D[MAXNODE], R[MAXNODE], L[MAXNODE], row[MAXNODE], col[MAXNODE]; int H[MAXN], S[MAXM]; int ansd, ans[MAXN]; void init(int n,int m) { this->n = n; this->m = m; ansd = INF; for(int i = 0; i <= m; i++) { S[i] = 0; U[i] = D[i] = i; L[i] = i - 1; R[i] = i + 1; } R[m] = 0; L[0] = m; size = m; for(int i = 1; i <= n; i++) H[i] = -1; } void Link(int r,int c) { ++S[col[++size] = c]; row[size] = r; D[size] = D[c]; U[D[c]] = size; U[size] = c; D[c] = size; if(H[r] < 0) H[r] = L[size] = R[size] = size; else { R[size] = R[H[r]]; L[R[H[r]]] = size; L[size] = H[r]; R[H[r]] = size; } } void remove(int c) { for(int i = D[c]; i != c; i = D[i]) { L[R[i]] = L[i]; R[L[i]] = R[i]; } } void resume(int c) { for(int i = U[c]; i != c; i = U[i]) L[R[i]] = R[L[i]] = i; } bool v[MAXNODE]; int f() { int ret = 0; for(int c = R[0]; c != 0; c = R[c]) v[c] = true; for(int c = R[0]; c != 0; c = R[c]) { if(v[c]) { ret++; v[c] = false; for(int i = D[c]; i != c; i = D[i]) for(int j = R[i]; j != i; j = R[j]) v[col[j]] = false; } } return ret; } void Dance(int d) { if(d + f() >= ansd) return; if(R[0] == 0) { if (d < ansd) ansd = d; return; } int c = R[0]; for(int i = R[0]; i != 0; i = R[i]) { if(S[i] < S[c]) c = i; } for(int i = D[c]; i != c; i = D[i]) { remove(i); for(int j = R[i]; j != i; j = R[j]) remove(j); ans[d] = row[i]; Dance(d + 1); for(int j = L[i]; j != i; j = L[j]) resume(j); resume(i); } } } gao; const int N = 20; int n, m, g[N][N], n1, m1, to[N][N]; int main() { while (~scanf("%d%d", &n, &m)) { int cnt = 0; for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) { scanf("%d", &g[i][j]); if (g[i][j]) to[i][j] = ++cnt; } scanf("%d%d", &n1, &m1); gao.init(n * m, cnt); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { for (int x = 0; x < n1 && i + x < n; x++) { for (int y = 0; y < m1 && j + y < m; y++) { if (g[i + x][j + y]) gao.Link(i * m + j + 1, to[i + x][j + y]); } } } } gao.Dance(0); printf("%d\n", gao.ansd); } return 0; }