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;
}