FZU 1686 神龙的难题（DLX反复覆盖）

FZU 1686 神龙的难题

#include <cstdio>
#include <cstring>

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

, n1, m1, to

;

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;
}