POJ 1739 Tony's Tour（插头DP）

Tony's Tour

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 2648		Accepted: 1197

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

 

 

插头DP。题目意思就是从左下角走到右下角，每个非障碍格子都走一遍的方法数。

一种方法是后面加两行转化成回路问题。

转成回路问题就和这题一样了：http://www.cnblogs.com/kuangbin/archive/2012/09/29/2708989.html

 

也可以不增加行，只要在起点和终点特殊处理下即可。

具体看两个代码：

/*
POJ 1739
题目意思就是从左下角走到右下角，每个非障碍格子都走一遍的方法数
转换成回路问题。
在最后加两行
   .########.
   ..........
这样就转成回路问题了，就和URAL 1519 一样的做法了

G++ 47ms
*/
#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;

const int MAXD=15;
const int HASH=10007;
const int STATE=1000010;

int N,M;
int maze[MAXD][MAXD];
int code[MAXD];
int ch[MAXD];//最小表示法使用
int ex,ey;//最后一个非障碍格子的坐标

struct HASHMAP
{
    int head[HASH],next[STATE],size;
    long long state[STATE],f[STATE];
    void init()
    {
        size=0;
        memset(head,-1,sizeof(head));
    }
    void push(long long st,long long ans)
    {
        int i;
        int h=st%HASH;
        for(i=head[h];i!=-1;i=next[i])
          if(state[i]==st)
          {
              f[i]+=ans;
              return;
          }
        state[size]=st;
        f[size]=ans;
        next[size]=head[h];
        head[h]=size++;
    }
}hm[2];

void decode(int *code,int m,long long st)
{
    for(int i=m;i>=0;i--)
    {
        code[i]=st&7;
        st>>=3;
    }
}
long long encode(int *code,int m)
{
    int cnt=1;
    memset(ch,-1,sizeof(ch));
    ch[0]=0;
    long long st=0;
    for(int i=0;i<=m;i++)
    {
        if(ch[code[i]]==-1)ch[code[i]]=cnt++;
        code[i]=ch[code[i]];
        st<<=3;
        st|=code[i];
    }
    return st;
}
void shift(int *code,int m)
{
    for(int i=m;i>0;i--)code[i]=code[i-1];
    code[0]=0;
}

void dpblank(int i,int j,int cur)
{
    int k,left,up;
    for(k=0;k<hm[cur].size;k++)
    {
        decode(code,M,hm[cur].state[k]);
        left=code[j-1];
        up=code[j];
        if(left&&up)
        {
            if(left==up)
            {
                if(i==ex&&j==ey)//只能出现在最后一个非障碍格子
                {
                    code[j]=code[j-1]=0;
                    if(j==M)shift(code,M);
                    hm[cur^1].push(encode(code,M),hm[cur].f[k]);
                }
            }
            else //不在同一个连通分量则合并
            {
                code[j-1]=code[j]=0;
                for(int t=0;t<=M;t++)
                  if(code[t]==left)
                    code[t]=up;
                if(j==M)shift(code,M);
                hm[cur^1].push(encode(code,M),hm[cur].f[k]);
            }
        }
        else if((left&&(!up))||((!left)&&up))
        {
            int t;
            if(left)t=left;
            else t=up;
            if(maze[i][j+1])
            {
                code[j-1]=0;
                code[j]=t;
                hm[cur^1].push(encode(code,M),hm[cur].f[k]);
            }
            if(maze[i+1][j])
            {
                code[j-1]=t;
                code[j]=0;
                if(j==M)shift(code,M);
                hm[cur^1].push(encode(code,M),hm[cur].f[k]);
            }
        }
        else
        {
            if(maze[i][j+1]&&maze[i+1][j])
            {
                code[j-1]=code[j]=13;
                hm[cur^1].push(encode(code,M),hm[cur].f[k]);
            }
        }
    }
}
void dpblock(int i,int j,int cur)
{
    int k;
    for(k=0;k<hm[cur].size;k++)
    {
        decode(code,M,hm[cur].state[k]);
        code[j-1]=code[j]=0;
        if(j==M)shift(code,M);
        hm[cur^1].push(encode(code,M),hm[cur].f[k]);
    }
}
char str[20];
void init()
{
    memset(maze,0,sizeof(maze));
    for(int i=1;i<=N;i++)
    {
        scanf("%s",&str);
        for(int j=0;j<M;j++)
          if(str[j]=='.')
          {
              maze[i][j+1]=1;
          }
    }
    maze[N+1][1]=maze[N+1][M]=1;
    for(int i=2;i<M;i++)maze[N+1][i]=0;
    for(int i=1;i<=M;i++)maze[N+2][i]=1;
    N+=2;
    ex=N,ey=M;
}
void solve()
{
    int i,j,cur=0;
    hm[cur].init();
    hm[cur].push(0,1);
    for(i=1;i<=N;i++)
      for(j=1;j<=M;j++)
      {
          hm[cur^1].init();
          if(maze[i][j])dpblank(i,j,cur);
          else dpblock(i,j,cur);
          cur^=1;
      }
    long long ans=0;
    for(int i=0;i<hm[cur].size;i++)
      ans+=hm[cur].f[i];
    printf("%I64d\n",ans);
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%d%d",&N,&M))
    {
        if(N==0&&M==0)break;
        init();
        solve();
    }
    return 0;
}

 

 

/*
POJ 1739
不增加行。
起点和终点特殊处理

G++ 47ms
*/
#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;

const int MAXD=15;
const int HASH=10007;
const int STATE=1000010;

int N,M;
int maze[MAXD][MAXD];
int code[MAXD];
int ch[MAXD];//最小表示法使用
int ex,ey;//最后一个非障碍格子的坐标

struct HASHMAP
{
    int head[HASH],next[STATE],size;
    long long state[STATE],f[STATE];
    void init()
    {
        size=0;
        memset(head,-1,sizeof(head));
    }
    void push(long long st,long long ans)
    {
        int i;
        int h=st%HASH;
        for(i=head[h];i!=-1;i=next[i])
          if(state[i]==st)
          {
              f[i]+=ans;
              return;
          }
        state[size]=st;
        f[size]=ans;
        next[size]=head[h];
        head[h]=size++;
    }
}hm[2];

void decode(int *code,int m,long long st)
{
    for(int i=m;i>=0;i--)
    {
        code[i]=st&7;
        st>>=3;
    }
}
long long encode(int *code,int m)
{
    int cnt=1;
    memset(ch,-1,sizeof(ch));
    ch[0]=0;
    long long st=0;
    for(int i=0;i<=m;i++)
    {
        if(ch[code[i]]==-1)ch[code[i]]=cnt++;
        code[i]=ch[code[i]];
        st<<=3;
        st|=code[i];
    }
    return st;
}
void shift(int *code,int m)
{
    for(int i=m;i>0;i--)code[i]=code[i-1];
    code[0]=0;
}

void dpblank(int i,int j,int cur)
{
    int k,left,up;
    for(k=0;k<hm[cur].size;k++)
    {
        decode(code,M,hm[cur].state[k]);
        left=code[j-1];
        up=code[j];

        if((i==N&&j==1)||(i==N&&j==M))//起点和终点
        {
            if((left&&(!up))||((!left)&&up))
            {
                code[j]=code[j-1]=0;
                if(j==M)shift(code,M);
                hm[cur^1].push(encode(code,M),hm[cur].f[k]);
            }
            else if(left==0&&up==0)
            {
                if(maze[i][j+1])
                {
                   code[j-1]=0;
                   code[j]=13;
                   hm[cur^1].push(encode(code,M),hm[cur].f[k]);
                }
                if(maze[i+1][j])
                {
                   code[j-1]=13;
                   code[j]=0;
                   if(j==M)shift(code,M);
                   hm[cur^1].push(encode(code,M),hm[cur].f[k]);
                }
            }
            continue;
        }
        if(left&&up)
        {
            if(left==up)
            {//这种情况不能发生
               /* if(i==ex&&j==ey)//只能出现在最后一个非障碍格子
                {
                    code[j]=code[j-1]=0;
                    if(j==M)shift(code,M);
                    hm[cur^1].push(encode(code,M),hm[cur].f[k]);
                }*/
            }
            else //不在同一个连通分量则合并
            {
                code[j-1]=code[j]=0;
                for(int t=0;t<=M;t++)
                  if(code[t]==left)
                    code[t]=up;
                if(j==M)shift(code,M);
                hm[cur^1].push(encode(code,M),hm[cur].f[k]);
            }
        }
        else if((left&&(!up))||((!left)&&up))
        {
            int t;
            if(left)t=left;
            else t=up;
            if(maze[i][j+1])
            {
                code[j-1]=0;
                code[j]=t;
                hm[cur^1].push(encode(code,M),hm[cur].f[k]);
            }
            if(maze[i+1][j])
            {
                code[j-1]=t;
                code[j]=0;
                if(j==M)shift(code,M);
                hm[cur^1].push(encode(code,M),hm[cur].f[k]);
            }
        }
        else
        {
            if(maze[i][j+1]&&maze[i+1][j])
            {
                code[j-1]=code[j]=13;
                hm[cur^1].push(encode(code,M),hm[cur].f[k]);
            }
        }
    }
}
void dpblock(int i,int j,int cur)
{
    int k;
    for(k=0;k<hm[cur].size;k++)
    {
        decode(code,M,hm[cur].state[k]);
        code[j-1]=code[j]=0;
        if(j==M)shift(code,M);
        hm[cur^1].push(encode(code,M),hm[cur].f[k]);
    }
}
char str[20];
void init()
{
    memset(maze,0,sizeof(maze));
    for(int i=1;i<=N;i++)
    {
        scanf("%s",&str);
        for(int j=0;j<M;j++)
          if(str[j]=='.')
          {
              maze[i][j+1]=1;
          }
    }
}
void solve()
{
    int i,j,cur=0;
    hm[cur].init();
    hm[cur].push(0,1);
    for(i=1;i<=N;i++)
      for(j=1;j<=M;j++)
      {
          hm[cur^1].init();
          if(maze[i][j])dpblank(i,j,cur);
          else dpblock(i,j,cur);
          cur^=1;
      }
    long long ans=0;
    for(int i=0;i<hm[cur].size;i++)
      ans+=hm[cur].f[i];
    printf("%I64d\n",ans);
}
int main()
{
    //freopen("in.txt","r",stdin);
   // freopen("out.txt","w",stdout);
    while(scanf("%d%d",&N,&M))
    {
        if(N==0&&M==0)break;
        init();
        solve();
    }
    return 0;
}