BZOJ3723 : PA2014Final Gra w podwajanie

暴力搜索出所有可行的形状，可以发现本质不同的形状数只有6000个左右。

对于每个形状，它的大小不超过$8\times 8$，故可以按照右下角为原点重建坐标系，用一个unsigned long long来存储。

然后对于每个中心，先进行第一步扩展，若能成功扩展，则扫描所有形状，看看是否匹配即可。

时间复杂度$O(6000nm)$。

#include<cstdio>
#include<algorithm>
using namespace std;
typedef unsigned long long ll;
const int N=210,M=30000;
int dx[4]={-1,1,0,0},dy[4]={0,0,-1,1};ll state[8][8];
int n,m,i,j,k,ans[6]={0,1,2,4,8,16};char a

,f

;
struct Shape{
int st
,en
,ed,v
[2],cnt[5][5];ll q[5][5][M];
inline void add(int x,int y){st[++ed]=x;en[ed]=y;}
void dfs(int x){
if(x>ed){
int a=0,b=0;
for(int i=1;i<=x;i++)a=max(a,v[i][0]),b=max(b,v[i][1]);
ll t=0;
for(int i=1;i<=x;i++)t|=state[a-v[i][0]][b-v[i][1]];
q[a][b][++cnt[a][b]]=t;
return;
}
for(int i=0;i<4;i++){
int X=v[st[x]][0]+dx[i],Y=v[st[x]][1]+dy[i],flag=1;
for(int j=1;j<=x;j++)if(v[j][0]==X&&v[j][1]==Y){flag=0;break;}
if(flag)v[en[x]][0]=X,v[en[x]][1]=Y,dfs(x+1);
}
}
void init(int k){
if(k==2){
add(1,2);
}
if(k==3){
add(1,2);
add(1,3);
add(2,4);
}
if(k==4){
add(1,2);
add(1,3);
add(1,4);
add(2,5);
add(2,6);
add(3,7);
add(5,8);
}
if(k==5){
add(1,2);
add(1,3);
add(1,4);
add(1,5);
add(2,6);
add(2,7);
add(2,8);
add(3,9);
add(3,10);
add(4,11);
add(6,12);
add(6,13);
add(7,14);
add(9,15);
add(12,16);
}
dfs(1);
for(int i=0;i<5;i++)for(int j=0;j<5;j++)if(cnt[i][j]){
sort(q[i][j]+1,q[i][j]+cnt[i][j]+1);
int k=0;
for(int x=1;x<=cnt[i][j];x++)if(q[i][j][x]!=q[i][j][x-1])q[i][j][++k]=q[i][j][x];
cnt[i][j]=k;
}
}
}G[6];
inline bool check(int x,int y,int z){
if(f[x][y]<z-1)return 0;
if(f[x][y]>=z)return 1;
if(z<5)return 1;
if(check(x-1,y,1)&&check(x+1,y,2)&&check(x,y-1,3)&&check(x,y+1,4))return 1;
if(check(x-1,y,1)&&check(x+1,y,2)&&check(x,y-1,4)&&check(x,y+1,3))return 1;
if(check(x-1,y,1)&&check(x+1,y,3)&&check(x,y-1,2)&&check(x,y+1,4))return 1;
if(check(x-1,y,1)&&check(x+1,y,3)&&check(x,y-1,4)&&check(x,y+1,2))return 1;
if(check(x-1,y,1)&&check(x+1,y,4)&&check(x,y-1,2)&&check(x,y+1,3))return 1;
if(check(x-1,y,1)&&check(x+1,y,4)&&check(x,y-1,3)&&check(x,y+1,2))return 1;
if(check(x-1,y,2)&&check(x+1,y,1)&&check(x,y-1,3)&&check(x,y+1,4))return 1;
if(check(x-1,y,2)&&check(x+1,y,1)&&check(x,y-1,4)&&check(x,y+1,3))return 1;
if(check(x-1,y,2)&&check(x+1,y,3)&&check(x,y-1,1)&&check(x,y+1,4))return 1;
if(check(x-1,y,2)&&check(x+1,y,3)&&check(x,y-1,4)&&check(x,y+1,1))return 1;
if(check(x-1,y,2)&&check(x+1,y,4)&&check(x,y-1,1)&&check(x,y+1,3))return 1;
if(check(x-1,y,2)&&check(x+1,y,4)&&check(x,y-1,3)&&check(x,y+1,1))return 1;
if(check(x-1,y,3)&&check(x+1,y,1)&&check(x,y-1,2)&&check(x,y+1,4))return 1;
if(check(x-1,y,3)&&check(x+1,y,1)&&check(x,y-1,4)&&check(x,y+1,2))return 1;
if(check(x-1,y,3)&&check(x+1,y,2)&&check(x,y-1,1)&&check(x,y+1,4))return 1;
if(check(x-1,y,3)&&check(x+1,y,2)&&check(x,y-1,4)&&check(x,y+1,1))return 1;
if(check(x-1,y,3)&&check(x+1,y,4)&&check(x,y-1,1)&&check(x,y+1,2))return 1;
if(check(x-1,y,3)&&check(x+1,y,4)&&check(x,y-1,2)&&check(x,y+1,1))return 1;
if(check(x-1,y,4)&&check(x+1,y,1)&&check(x,y-1,2)&&check(x,y+1,3))return 1;
if(check(x-1,y,4)&&check(x+1,y,1)&&check(x,y-1,3)&&check(x,y+1,2))return 1;
if(check(x-1,y,4)&&check(x+1,y,2)&&check(x,y-1,1)&&check(x,y+1,3))return 1;
if(check(x-1,y,4)&&check(x+1,y,2)&&check(x,y-1,3)&&check(x,y+1,1))return 1;
if(check(x-1,y,4)&&check(x+1,y,3)&&check(x,y-1,1)&&check(x,y+1,2))return 1;
if(check(x-1,y,4)&&check(x+1,y,3)&&check(x,y-1,2)&&check(x,y+1,1))return 1;
return 0;
}
inline bool ok(int x,int y){
if(x<1||x>n||y<1||y>m)return 0;
return a[x][y];
}
inline bool judge(int x,int y,int z){
int i,j,k,l;
for(i=0;i<5;i++)for(j=0;j<5;j++){
ll t=0;
for(k=0;k<8;k++)for(l=0;l<8;l++)if(ok(x+i-k,y+j-l))t|=state[k][l];
for(k=G[z].cnt[i][j];k;k--)if((G[z].q[i][j][k]&t)==G[z].q[i][j][k])return 1;
}
return 0;
}
int main(){
for(i=0;i<8;i++)for(j=0;j<8;j++)state[i][j]=1ULL<<(i*8+j);
for(i=2;i<=5;i++)G[i].init(i);
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)for(scanf("%s",a[i]+1),j=1;j<=m;j++)a[i][j]-='0';
for(i=1;i<=n;i++)for(j=1;j<=m;j++)f[i][j]=a[i][j];
for(k=2;k<=5;k++)for(i=1;i<=n;i++)for(j=1;j<=m;j++)if(check(i,j,k))if(judge(i,j,k))f[i][j]++;
for(i=1;i<=n;i++)for(j=1;j<=m;j++)printf("%d%c",ans[f[i][j]],j<m?' ':'\n');
return 0;
}