POJ 1739 Tony's Tour 楼教主的男人八题之一
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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 Source
LouTiancheng@POJ
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这题和Ural 1519 Formula 1类似。只是把回路换成以左下角作为起点。右下角换成终点的回路数而已。所以预处理加两行就行。
比如：
3 4 . . . .
. . . . . . . .
. . . . ----------> . . . .
. . . . . ##.
. . . .
这样预处理后就和那题一样了。
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int HASH=30007;//哈希表的大小
const int STATE=1000010;//状态数
const int MAXD=15;
int N,M,ex,ey;//ex,ey最后一个可到点
int code[MAXD],maze[MAXD][MAXD],mcod[MAXD];//编码。和存地图。最小表示法
char s[20];
struct HASHMAP//哈希表结构
{
int head[HASH],next[STATE],sz;//哈希表头指针模数相同的状态用链表连接。方便状态查找和判重
long long f[STATE],state[STATE];//f记录对应状态的方法数。next指向模数相同的下一个状态。state记录状态。sz记录状态总数
void init()//哈希表初始化函数
{
sz=0;
memset(head,-1,sizeof(head));
}
void push(long long st,long long ans)//压入状态和方法数
{
int i,h=st%HASH;
for(i=head[h]; i!=-1; i=next[i])
if(st==state[i])//若状态已经存在。方法数增加就行
{
f[i]+=ans;
return;
}
f[sz]=ans;//存入新的可行状态
state[sz]=st;
next[sz]=head[h];
head[h]=sz++;
}
} hm[2];
void decode(int *code,int m,long long st)//编码从高位到低位m到0编码。对应从左到右的插头
{
int i;
for(i=m; i>=0; i--)
{
code[i]=st&7;
st>>=3;
}
}
long long encode(int *code,int m)//最小表示法解码到st中
{
int i,cnt=1;
long long st=0;
memset(mcod,-1,sizeof mcod);
mcod[0]=0;
for( i=0; i<=m; i++)
{
if(mcod[code[i]]==-1)
mcod[code[i]]=cnt++;
code[i]=mcod[code[i]];
st<<=3;
st|=code[i];
}
return st;
}
void init()//读数据。初始化
{
int i,j;
memset(maze,0,sizeof maze);
ex=0;
for(i=1; i<=N; i++)
{
scanf("%s",s+1);
for( j=1; j<=M; j++)
if(s[j]=='.')
maze[i][j]=1;
}
maze[N+1][1]=maze[N+1][M]=1;
for(i=2;i<M;i++)
maze[N+1][i]=0;
for(i=1;i<=M;i++)
maze[N+2][i]=1;
N+=2;
ex=N,ey=M;
}
void shift(int *code,int m)//换行的时候移位
{
int i;
for(i=m; i>0; i--)
code[i]=code[i-1];
code[0]=0;
}
void dpblank(int i,int j,int cur)//处理可到格的情况
{
int k,t,left,up,temp;
for(k=0; k<hm[cur].sz; k++) //遍历j格出轮廓线的状态进行状态转移
{
decode(code,M,hm[cur].state[k]);//对状态进行编码
left=code[j-1];//获取左插头状态
up=code[j];//获取上插头状态
if(left&&up)//11 -> 00
{
if(left==up)//形成回路
{
if(i==ex&&j==ey)//判断是否到了最后一个可到点
{
code[j-1]=code[j]=0;//只能为0一个格子只能两个插头
if(j==M)shift(code,M);//到了边界后换行
hm[cur^1].push(encode(code,M),hm[cur].f[k]);//压入新状态
}
}
else
{
code[j-1]=code[j]=0;
for(t=0;t<=M;t++)
if(code[t]==up)
{
code[t]=left;
break;
}
if(j==M)shift(code,M);//到了边界后换行
hm[cur^1].push(encode(code,M),hm[cur].f[k]);//压入新状态
}
}
else if((!left&&up)||(left&&!up))//01 或 10
{
if(up)
temp=up;
else
temp=left;
if(maze[i][j+1])//j==m时maze[i][j]为0也不用shift
{
code[j-1]=0;
code[j]=temp;
hm[cur^1].push(encode(code,M),hm[cur].f[k]);
}
if(maze[i+1][j])
{
code[j-1]=temp;
code[j]=0;
if(j==M)shift(code,M);//这个不要忘了！
hm[cur^1].push(encode(code,M),hm[cur].f[k]);
}
}
else
{
if(maze[i][j+1]&&maze[i+1][j])//若j==m不会合法所以不用shift
{
code[j]=code[j-1]=13;
hm[cur^1].push(encode(code,M),hm[cur].f[k]);
}
}
}
}
void dpblock(int i,int j,int cur)//处理不能到格的情况
{
int k;
for(k=0; k<hm[cur].sz; k++) //存入状态均合法不用判断
{
decode(code,M,hm[cur].state[k]);//先编码
code[j-1]=code[j]=0;//肯定不能用插头
if(j==M)shift(code,M);//转移到下一行
hm[cur^1].push(encode(code,M),hm[cur].f[k]);
}
}
void solve()
{
int i,j,cur=0;
long long ans=0;
hm[cur].init();//cur用于滚动数组。节约空间。cur存当前格上插头和左插头情况。cur^1用于记录转移出的新状态即下一格
hm[cur].push(0,1);
for(i=1; i<=N; i++)
for(j=1; j<=M; j++)
{
hm[cur^1].init();//初始化
if(maze[i][j])dpblank(i,j,cur);
else dpblock(i,j,cur);
cur^=1;
}
for(i=0; i<hm[cur].sz; i++)
ans+=hm[cur].f[i];
printf("%I64d\n",ans);
}
int main()
{
while(scanf("%d%d",&N,&M),N||M)
{
init();
if(!ex)
printf("0\n");
else
solve();
}
return 0;
}