poj2032Square Carpets(IDA* + dancing links)

题目请戳这里

题目大意:给一个H行W列的01矩阵,求最少用多少个正方形框住所有的1.

题目分析:又是一个红果果的重复覆盖模型.DLX搞之!

枚举矩阵所有的子正方形,全1的话建图.判断全1的时候,用了一个递推,dp[i][j][w][h]表示左上角(i,j)的位置开始长h宽w的矩形中1的个数,这样后面可以迅速判断某个正方形是否全1.

不过此题直接搜一直TLE,然后改成迭代加深就比较愉快啦

详情请见代码:

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 11;
const int M = 50005;
int dp[N][N][N][N];
int n,m,num,ans;
int mp[N][N];
bool vis[105];
int s[M],h[M],col[M],u[M],d[M],l[M],r[M];

void init()
{
    memset(h,0,sizeof(h));
    memset(s,0,sizeof(s));
    int i,c;
    c = m * n;
    for(i = 0;i <= c;i ++)
    {
        u[i] = d[i] = i;
        l[i] = (i + c)%(c + 1);
        r[i] = (i + 1)%(c + 1);
    }
    num = c + 1;
}

void ins(int i,int j)
{
    if(h[i])
    {
        r[num] = h[i];
        l[num] = l[h[i]];
        r[l[num]] = l[r[num]] = num;
    }
    else
        h[i] = l[num] = r[num] = num;
    s[j] ++;
    u[num] = u[j];
    d[num] = j;
    d[u[num]] = num;
    u[j] = num;
    col[num] = j;
    num ++;
}
void del(int c)
{
    for(int i = u[c];i != c;i = u[i])
        l[r[i]] = l[i],r[l[i]] = r[i];
}
void resume(int c)
{
    for(int i = d[c];i != c;i = d[i])
        l[r[i]] = r[l[i]] = i;
}

int A()
{
    int i,j,k,ret;
    ret = 0;
    memset(vis,false,sizeof(vis));
    for(i = r[0];i;i = r[i])
    {
        if(vis[i] == false)
        {
            ret ++;
            vis[i] = true;
            for(j = d[i];j != i;j = d[j])
                for(k = r[j];k != j;k = r[k])
                    vis[col[k]] = true;
        }
    }
    return ret;
}

bool dfs(int k,int lim)
{
    if(k + A() > lim)
        return false;
    if(!r[0])
    {
//        ans = min(ans,k);
        return true;
    }
    int i,j,c,mn = M;
    for(i = r[0];i;i = r[i])
    {
        if(s[i] < mn)
        {
            mn = s[i];
            c = i;
        }
    }
    for(i = d[c];i != c;i = d[i])
    {
        del(i);
        for(j = l[i];j != i;j = l[j])
            del(j);
        if(dfs(k + 1,lim)) return true;
        for(j = r[i];j != i;j = r[j])
            resume(j);
        resume(i);
    }
    return false;
}

bool canfuck(int x,int y,int z)
{
    return dp[x][y][z][z] == z * z;
    int i,j;
    for(i = x;i <= x + z - 1;i ++)
        for(j = y;j <= y + z - 1;j ++)
            if(mp[i][j] == 0)
                return false;
    return true;
}

void fuckba(int x,int y,int z,int id)
{
    int i,j;
    for(i = x;i <= x + z - 1;i ++)
        for(j = y;j <= y + z - 1;j ++)
            ins(id,(i - 1) * m + j);
}
void build()
{
    int i,j,k;
    memset(dp,0,sizeof(dp));
    for(i = 1;i <= n;i ++)
        for(j = 1;j <= m;j ++)
            scanf("%d",&mp[i][j]),dp[i][j][1][1] = mp[i][j];
    for(int ii = 1;ii <= n;ii ++)
        for(int jj = 1;jj <= m;jj ++)
        {
            for(i = 1;i + ii <= n + 1;i ++)
            {
                for(j = 1;j + jj <= 1 + m;j ++)
                {
                    dp[i][j][ii][jj] = dp[i][j][ii - 1][jj - 1] + dp[i + ii - 1][j][1][jj - 1] + dp[i][j + jj - 1][ii - 1][1] + dp[i + ii - 1][j + jj - 1][1][1];
                }
            }
        }
    init();
    int rownum = 1;
    for(k = 1;k <= min(n,m);k ++)//枚举边长
    {
        for(i = 1;i <= n - k + 1;i ++)
        {
            for(j = 1;j <= m - k + 1;j ++)
            {
                if(canfuck(i,j,k))
                    fuckba(i,j,k,rownum);
                rownum ++;
            }
        }
    }
    for(i = 1;i <= n * m;i ++)
        if(s[i] == 0)
            l[r[i]] = l[i],r[l[i]] = r[i];
}
void fuck()
{
//    ans = M;TLE....
//    dfs(0);
//    printf("%d
",ans);
    ans = 0;
    while(!dfs(0,ans ++));
    printf("%d
",ans - 1);
}
int main()
{
    while(scanf("%d%d",&m,&n),(m + n))
    {
        build();
        fuck();
    }
    return 0;
}