FZU 1686 神龙的难题 (DLX求重复覆盖)

题目大意：

就是现在给出一个n*m的01矩阵

每次可以选择将一个n1*m1的矩形中的所有1变成0

问将所有的1变成0需要使用n1*m1的矩形至少多少次

大致思路：

对于每一个位置作为列, 一共15*15列, 对于每一个可能选择的矩形都建立一行, 然后用Dancing Link做重复覆盖即可

代码如下：

Result  :  Accepted     Memory  :  2348 KB     Time  :  62 ms

/*
* Author: Gatevin
* Created Time: 2015/10/4 18:10:46
* File Name: Sakura_Chiyo.cpp
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;

#define maxnode 90000
#define maxn 300
#define maxm 300

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)
{
n = _n;
m = _m;
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)//因为是重复覆盖每次讲某一列的所有1纳入不考虑范围
{
//L[R[c]] = L[c]; R[L[c]] = R[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;
//L[R[c]] = R[L[c]] = c;
}
bool v[60];
int f()//A*搜索估价函数
{
int ret = 0;
for(int i = R[0]; i != 0; i = R[i]) v[i] = 1;
for(int i = R[0]; i != 0; i = R[i])
if(v[i])
{
ret++;
v[i] = 0;
for(int j = D[i]; j != i; j = D[j])
for(int k = R[j]; k != j; k = R[k])
v[Col[k]] = 0;
}
return ret;
}
int res;
void Dance(int dep)
{
if(dep + f() >= res) return;
if(R[0] == 0)
{
res = dep;
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])
{
for(int j = R[i]; j != i; j = R[j]) remove(j);
remove(i);
Dance(dep + 1);
resume(i);
for(int j = L[i]; j != i; j = L[j]) resume(j);
}
return;
}
int maz[20][20];
int col[20][20];
void solve()
{
int N, M;
while(~scanf("%d %d", &N, &M))
{
int cnt = 0;
for(int i = 1; i <= N; i++)
for(int j = 1; j <= M; j++)
{
scanf("%d", &maz[i][j]);
if(maz[i][j] == 1) col[i][j] = ++cnt;
}
int n1, m1;
scanf("%d %d", &n1, &m1);
init((N - n1 + 1)*(M - m1 + 1), cnt);
for(int i = 1; i + m1 - 1 <= M; i++)
for(int j = 1; j + n1 - 1 <= N; j++)
{
int row = (j - 1)*(M - m1 + 1) + i;
for(int x = 1; x <= m1; x++)
for(int y = 1; y <= n1; y++)
{
int px = i + x - 1;
int py = j + y - 1;
if(maz[py][px] == 1)
Link(row, col[py][px]);
}
}
res = (N - n1 + 1)*(M - m1 + 1);
Dance(0);
printf("%d\n", res);
}
return;
}
};

DLX dlx;

int main()
{
dlx.solve();
return 0;
}