fzu1868重复覆盖模板

最新推荐文章于 2018-07-25 15:38:39 发布

原创最新推荐文章于 2018-07-25 15:38:39 发布 · 200 阅读

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本文介绍了一种使用DLX算法解决重复覆盖模板题的方法。通过定义结构体DLX，实现了包括初始化、链接、移除和恢复等核心操作。针对具体问题，如给定矩阵中寻找最优覆盖方案，该算法能有效求解。

重复覆盖模板题。

1为列，可以操作的为行

#include<cstdio>
#include<cstring>
using namespace std;
int num[20][20];
int bmap[20][20];
const int MaxM = 15*15+10;
const int MaxN = 15*15+10;
const int maxnode = MaxN * MaxM;
const int INF = 0x3f3f3f3f;
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;
    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)
    {
        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[MaxM];
    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);
            Dance(d+1);
            for(int j = L[i];j != i;j = L[j])resume(j);
            resume(i);
        }
    }
}dlx;
int main()
{
    int n,m,a,b;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(bmap,0,sizeof(bmap));
        int sz = 0;
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++){
                scanf("%d",&num[i][j]);
                if(num[i][j]==1) bmap[i][j] = ++sz;
            }
        dlx.init(n*m,sz);
        scanf("%d%d",&a,&b); int tot = 1;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                for(int x=0;x<a&&i+x<n;x++)
                    for(int y =0;y<b&&j+y<m;y++)
                    if(bmap[i+x][j+y])
                    dlx.Link(tot,bmap[i+x][j+y]);
                tot++;
            }
        }
        dlx.ansd = n*m+1;
        dlx.Dance(0);
        printf("%d\n",dlx.ansd);
    }
    return 0;
}