Light OJ 1270 Tiles (II) (状压DP)

解析：

显然的状压DP，每一行怎么填补空缺用搜索就好，就是繁琐了些。

[code]:

#include<cstdio>
#include<cstring>
#include<algorithm>
 
using namespace std;
typedef unsigned long long LL;
 
char s[105][105];
int n,m,a[105],b[8],c[8],d;
LL dp[105][1<<8],tmp_val;
 
void preprocess(){
    int i,j;
    if(n > m){
        for(i = 0;i < n;i++){
            a[i] = 0;
            for(j = 0;j < m;j++){
                if(s[i][j]=='.') a[i] |= (1<<j);
            }
        }
    }else{
        for(j = 0;j < m;j++){
            a[j] = 0;
            for(i = 0;i < n;i++){
                if(s[i][j]=='.') a[j] |= (1<<i);
            }
        }
        swap(n,m);
    }
}
 
void dfs(int p,int i){
    int pos = b[p];
    if(pos == m){
        int j,S = 0;
        for(j = 0;j < m;j++) if(c[j]) S |= (1<<j);
        if(S|a[i]==a[i]){
            dp[i+1][S] += tmp_val;
//printf("--> %d %d %llu\n",i+1,S,dp[i+1][S]);
        }
        return;
    }
    if(c[pos]==0){
        c[pos] = 1;
        dfs(p+1,i);
        c[pos] = 0;
        if(pos>0&&c[pos-1]==0){
            c[pos-1] = c[pos] = 1;
            dfs(p+1,i);
            c[pos-1] = c[pos] = 0;
        }
        if(pos<m-1&&c[pos+1]==0){
            c[pos] = c[pos+1] = 1;
            dfs(p+1,i);
            c[pos] = c[pos+1] = 0;
        }
    }
    if(b[p+1]!=m&&b[p+1]==pos+1){
        dfs(p+2,i);
        if(c[pos]==0){
            c[pos] = 1;
            dfs(p+2,i);
            c[pos] = 0;
        }
        if(c[pos+1]==0){
            c[pos+1] = 1;
            dfs(p+2,i);
            c[pos+1] = 0;
        }
    }
}
 
void sol(){
//puts("");
    int i,j,S,S_,Ed = 1<<m;
    memset(dp,0,sizeof(dp));
    dp[0][0] = 1;
    for(i = 0;i < n-1;i++){
        for(S = 0;S < Ed;S++){
            if((S|a[i])!=a[i]||dp[i][S]==0) continue;
            tmp_val = dp[i][S];
            d = 0;
            for(j = 0;j < m;j++){
                if(!(S>>j&1)&&(a[i]|(1<<j))==a[i]){
                    b[d++] = j;
                }
                c[j] = 0;
            }
            b[d++] = m;
            dfs(0,i);
        }
//printf("%d -> ",i+1);
//for(S = 0;S < Ed;S++) printf(" %llu",dp[i+1][S]);putchar('\n');
    }
    LL ans = 0;
    for(S = 0;S < Ed;S++){
        S_ = S;
        for(i = 0;i < n;i++){
            if(!(S_>>i&1)&&!(S_>>(i+1)&1)&&(S_|(1<<i)|(1<<(i+1))|a[n-1])==a[n-1]){
                S_ |= (1<<i);
                S_ |= (1<<(i+1));
            }
        }
        if(S_==a[n-1]) ans += dp[n-1][S];
    }
    printf("%llu\n",ans);
}
 
int main(){
    int i,j,cas;
    scanf("%d",&cas);
    for(int T=1;T<=cas;T++){
        scanf("%d%d",&n,&m);
        for(i = 0;i < n;i++) scanf("%s",s[i]);
        preprocess();
        printf("Case %d: ",T);
        sol();
    }
 
    return 0;
}