poj 1739（男人八题之插头DP）

八题算法挑战：Tony'sTour路径计数

最新推荐文章于 2020-08-27 12:21:46 发布

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本文探讨了Tony'sTour问题，即在一个由n行m列的正方形地块组成的农场中，从左下角的农场出发，经过所有未被阻挡的地块恰好一次到达右下角市场，计算可能的不同路径数量。通过引入回路算法和类似Ural1519题目的解题思路，提供了一个简洁高效的解决方法。

Tony's Tour

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 1680		Accepted: 751

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

Source

LouTiancheng@POJ

分析：传说中的男人八题，哈哈，此题一看以为是插头DP简单路径问题，但是仔细想，由于起点和终点固定，所以完全可以用一条回路的算法，具体做法和ural 1519类似，直接套用那题的模板，稍微改一下结尾，哈哈。。。水过‍

代码：

#include<cstdio>
#include<cstring>
#define mm 1007
#define mn 11
struct hashtable
{
    int h[mm],s[mm],p[mm],sz;
    long long d[mm];
    int hash(int x,long long v)
    {
        int c=x%mm,i;
        for(i=h[c];i>=0;i=p[i])
        if(s[i]==x)
        {
            d[i]+=v;
            return i;
        }
        d[sz]=v,s[sz]=x,p[sz]=h[c],h[c]=sz;
        return sz++;
    }
    void clear()
    {
        sz=0,memset(h,-1,sizeof(h));
    }
}f[2];
bool g[15][15];
int g1,g2,i,j,k,x,y,s,n,m;
int eat(bool f)
{
    int a=x<<(j<<1),b=(3^x)<<((j+f)<<1),c=3<<((j+f)<<1),n=1;
    s=s^a^(a<<2);
    if(f)a<<=2;
    while(n)
    {
        if(f)a<<=2,b<<=2,c<<=2;
        else a>>=2,b>>=2,c>>=2;
        x=c&s;
        if(x==a)++n;
        if(x==b)--n;
    }
    return (s^b)|a;
}
inline bool ok(int c)
{
    if(c==1)return g[i+1][j];
    if(c==2)return g[i][j+1];
    if(c==3)return g[i+1][j]&&g[i][j+1];
    return 1;
}
void move(long long v)
{
    if(x==0&&y==0)
    {
        if(ok(3))f[g2].hash(s|9<<(j<<1),v);
    }
    else if(x==0||y==0)
    {
        if(x)y=1;else x=y,y=2;
        x=x<<(j<<1);
        if(ok(y))f[g2].hash(s,v);
        if(ok(3^y))f[g2].hash(s^x^(x<<2),v);
    }
    else if(x==y)f[g2].hash(eat(x==1),v);
    else if(x==2&&y==1)f[g2].hash(s^(6<<(j<<1)),v);
}
long long PlugDP()
{
    f[0].clear();
    f[0].hash(0,1);
    for(g1=1,g2=i=0;i<n;++i)
    {
        for(j=0;j<f[g2].sz;++j)f[g2].s[j]<<=2;
        for(j=0;j<m;++j)
            if(g[i][j])for(g1=!g1,g2=!g2,f[g2].clear(),k=0;k<f[g1].sz;++k)
            {
                s=f[g1].s[k],x=(s>>(j<<1))&3,y=(s>>((j+1)<<1))&3;
                move(f[g1].d[k]);
            }
    }
    long long ret=0;
    for(i=0;i<f[g2].sz;++i)
        if(f[g2].s[i]==(1|(2<<((m-1)<<1))))ret+=f[g2].d[i];
    return ret;
}
int main()
{
    char c;
    while(scanf("%d%d",&n,&m),n)
    {
        memset(g,0,sizeof(g));
        for(i=0;i<n;++i)
            for(j=0;j<m;++j)
                scanf(" %c",&c),g[i][j]=(c=='.');
        g[n][0]=g[n][m-1]=1;
        printf("%lld\n",PlugDP());
    }
    return 0;
}