POJ 1739：Tony's Tour

Description

A square township has been divided up into n*m(n rows and m columns) square plots (1<=N,M<=8),some of them are blocked, others are unblocked. The Farm is located in the lower left plot and the Market is located in the lower right plot. Tony takes her tour of the township going from Farm to Market by walking through every unblocked plot exactly once. 
Write a program that will count how many unique tours Betsy can take in going from Farm to Market. 

Input

The input contains several test cases. The first line of each test case contain two integer numbers n,m, denoting the number of rows and columns of the farm. The following n lines each contains m characters, describe the farm. A '#' means a blocked square, a '.' means a unblocked square. 
The last test case is followed by two zeros. 

Output

For each test case output the answer on a single line.

Sample Input

2 2
..
..
2 3
#..
...
3 4
....
....
....
0 0

Sample Output

1
1
4


楼教主八题之一，同样可以加行，也可以特殊处理，我写的是加行。

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mn=177147;
int i;
struct na{
    int x,z;
    na(int xx,int zz):x(xx),z(zz){}
};
int n,m,x,y,z,a[21],k,en,p1,p2;
char u[100];
bool map[21][21];
long long f[2][mn+1],ans;
int v[2][mn+1];
int re[21][21];
queue <na> q;
inline int gx(int x,int q1,int q2){k=0;for (register int i=m+1;i;i--) k=k*3+(i==x?q1:(i==x+1?q2:a[i]));return k;}
inline void up(int x,int z,long long lj){
    x++;
    k=x%2;
    if (v[k][z]!=x) v[k][z]=x,f[k][z]=0,q.push(na(x,z));
    f[k][z]+=lj;
}
int main(){
    register int i,j;
    while(scanf("%d%d",&n,&m)!=EOF){
        if (n==0&&m==0) break;
        ans=0;
        memset(map,0,sizeof(map));memset(v,0,sizeof(v));memset(re,0,sizeof(re));memset(f,0,sizeof(f));
        for (i=1;i<=n;i++){
            scanf("%s",&u);
            for (j=1;j<=m;j++)
            map[j][n-i+3]=u[j-1]=='.';
        }
        n+=2;
        en=n*m-1;
        map[1][2]=1;map[m][2]=1;
        for (i=1;i<=m;i++) map[i][1]=1;
        while(!map[en%m+1][en/m+1]) en--;
        f[0][0]=v[0][0]=1;
        q.push(na(0,0));
        while(!q.empty()){
            na no=q.front();q.pop();
            long long an=f[no.x%2][no.z];
            if(no.x%m==0) no.z*=3;
            x=no.x%m+1;y=no.x/m+1;
            for (i=1;i<=m+1;i++) a[i]=0;
            for (i=1,j=no.z;j;i++,j/=3) a[i]=j%3;
            if (!map[x][y])up(no.x,gx(x,0,0),an);else
            if (a[x]==1&&a[x+1]==2){
                if (no.x==en) ans+=an;
            }else if (a[x]==2&&a[x+1]==1) up(no.x,gx(x,0,0),an);else
                if (a[x]==0&&a[x+1]==0){
                if (map[x][y+1]&&map[x+1][y]) up(no.x,gx(x,1,2),an);
            }else if (a[x]==0){
                if (map[x+1][y]) up(no.x,gx(x,0,a[x+1]),an);
                if (map[x][y+1]) up(no.x,gx(x,a[x+1],0),an);
            }else if (a[x+1]==0){
                if (map[x+1][y]) up(no.x,gx(x,0,a[x]),an);
                if (map[x][y+1]) up(no.x,gx(x,a[x],0),an);
            }else if (a[x]==a[x+1]){
                p1=p2=0;
                if (a[x]==1)
                for (j=0,i=x+2;i<=m;i++){
                    if (a[i]==1) j--;
                    if (a[i]==2) j++;
                    if (j>0&&!p1) p1=i,j--;
                    if (j>0&&p1){p2=i;break;}
                }else
                for (j=0,i=x-1;i;i--){
                    if (a[i]==1) j++;
                    if (a[i]==2) j--;
                    if (j>0&&!p2) p2=i,j--;
                    if (j>0&&p2){p1=i;break;}
                }
                a[p1]=1;a[p2]=2;up(no.x,gx(x,0,0),an);
            }
        }
        printf("%I64d
",ans);
    }
}