poj2032Square Carpets(IDA* + dancing links)
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\n",ans);
ans = 0;
while(!dfs(0,ans ++));
printf("%d\n",ans - 1);
}
int main()
{
while(scanf("%d%d",&m,&n),(m + n))
{
build();
fuck();
}
return 0;
}