ZOJ 3256 Tour in the Castle (插头DP求回路个数+矩阵乘法）

Tour in the Castle

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Time Limit: 5 Seconds      Memory Limit: 32768 KB

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After the final BOSS is defeated, the hero found that the whole castle is collapsing (very familiar scene, isn't it). Escape from the castle is easy, just need to cross a few rooms. But as the Hero is full of adventurous spirit, he decides to visit every room before he escape the castle.

The castle is a rectangle with N * M rooms in it. Two rooms are connected if they share a common edge. The hero starts in the top left room. And the bottom left room is the only way out. After the hero visits a room and leaves it, it will collapse immediately(Another familiar scene). So he can visit each room only once.

The diagram shows one tour over a castle with 4 * 10 rooms:

Input

There are multiply cases (<20), process to the end of file.

Each case contains a line with two Integer N and M (2 <= N <= 7, 1 <= M <=10^9).

Ouput

For each case, if it's impossible to visit every room exactly once and get to the bottom left room, output "Impossible". Otherwise, output the number of tours as it describe above. Beacause the answer can be huge, you just need to output the answer MOD 7777777.

Sample Input

3 2

3 3

4 10

Sample Output

Impossible

2

2329

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Author: WANG, Yelei
Source: ZOJ Monthly, September 2009

 

 

初看题目。很熟悉的插头DP求环路个数。但是M很大，要用矩阵乘法。

按照列来DP。每一列的转移情况是相同的。

很好的题目。

但是因为在计算矩阵乘法的时候每次取模就超时了，加完之后取模就AC。坑啊。。感觉差不了多少呢。

 

/*

ZOJ 3256

N*M(2<=N<=7,1<=M<=10^9)的方格，问从左上角的格子到左下角的格子，

而且仅经过所有格子一次的路径数

插头DP+矩阵加速



对于一个图的邻接矩阵的N次方，其中(i,j)位置上的元素表示

点i经过N步到达点ｊ的方案数

*/

#include<stdio.h>

#include<iostream>

#include<algorithm>

#include<string.h>

using namespace std;



const int STATE=1010;

const int HASH=419;//这个小一点，效率高

const int MOD=7777777;



int N,M;

int D;

int code[10];

int ch[10];

int g[200][200];//状态转移图



struct Matrix

{

    int n,m;

    int mat[200][200];

};

Matrix mul(Matrix a,Matrix b)//矩阵相乘，要保证a的列数和b的行数相等

{

    Matrix ret;

    ret.n=a.n;

    ret.m=b.m;

    long long sum;

    for(int i=0;i<a.n;i++)

       for(int j=0;j<b.m;j++)

       {

           sum=0;

           for(int k=0;k<a.m;k++)

           {

               sum+=(long long)a.mat[i][k]*b.mat[k][j];

               //sum%=MOD;//加了这句话就会TLE，坑啊。。。

           }

           ret.mat[i][j]=sum%MOD;

       }

    return ret;

}

Matrix pow_M(Matrix a,int n)//方阵的n次方

{

    Matrix ret=a;

    memset(ret.mat,0,sizeof(ret.mat));

    for(int i=0;i<a.n;i++)ret.mat[i][i]=1;//单位阵

    Matrix temp=a;

    while(n)

    {

        if(n&1)ret=mul(ret,temp);

        temp=mul(temp,temp);

        n>>=1;

    }

    return ret;

}



struct HASHMAP

{

    int head[HASH],next[STATE],size;

    int state[STATE];

    void init()

    {

        size=0;

        memset(head,-1,sizeof(head));

    }

    int push(int st)

    {

        int i,h=st%HASH;

        for(i=head[h];i!=-1;i=next[i])

           if(state[i]==st)

              return i;

        state[size]=st;

        next[size]=head[h];

        head[h]=size++;

        return size-1;

    }

}hm;

void decode(int *code,int n,int st)

{

    for(int i=n-1;i>=0;i--)

    {

        code[i]=st&3;

        st>>=2;

    }

}

int encode(int *code,int n)

{

    int cnt=1;

    memset(ch,-1,sizeof(ch));

    ch[0]=0;

    int st=0;

    for(int i=0;i<n;i++)

    {

        if(ch[code[i]]==-1)ch[code[i]]=cnt++;

        code[i]=ch[code[i]];

        st<<=2;

        st|=code[i];

    }

    return st;

}



bool check(int st,int nst)//判断两种状态能不能转移

{

    decode(code,N,st);

    int flag=0;//标记格子上边是否有插头

    int cnt=0;

    int k;

    for(int i=0;i<N;i++)

    {

        if(flag==0)//这个格子上边没有插头

        {

            if(code[i]==0&&(nst&(1<<i))==0)//左边和右边都没有插头

               return false;

            if(code[i]&&(nst&(1<<i)))continue;

            if(code[i])flag=code[i];//插头从左边过来，从下边出去

            else flag=-1;//插头从下边进来从右边出去

            k=i;

        }

        else

        {

            if(code[i]&&(nst&(1<<i)))//左边和右边和上边都有插头

               return false;

            if(code[i]==0&&(nst&(1<<i))==0)continue;

            if(code[i])

            {

                if(code[i]==flag&&((nst!=0)||i!=N-1))return false;//只有最后一个格子才能合起来

                if(flag>0)

                {

                    for(int j=0;j<N;j++)

                      if(code[j]==code[i]&&j!=i)

                          code[j]=code[k];

                    code[i]=code[k]=0;

                }

                else

                {

                    code[k]=code[i];

                    code[i]=0;

                }

            }

            else

            {

                if(flag>0)code[i]=code[k],code[k]=0;

                else code[i]=code[k]=N+(cnt++);

            }

            flag=0;

        }

    }

    if(flag!=0)return false;

    return true;

}

struct Node

{

    int g[200][200];

    int D;

}node[20];//打表之用

void init()

{

    if(node[N].D!=0)

    {

        memcpy(g,node[N].g,sizeof(node[N].g));

        D=node[N].D;

        return;

    }

    int st,nst;

    hm.init();

    memset(code,0,sizeof(code));

    code[0]=code[N-1]=1;

    hm.push(0);

    hm.push(encode(code,N));

    memset(g,0,sizeof(g));

    for(int i=1;i<hm.size;i++)

    {

        st=hm.state[i];

        for(nst=0;nst<(1<<N);nst++)

          if(check(st,nst))

          {

              int j=hm.push(encode(code,N));

              g[i][j]=1;

          }

    }

    D=hm.size;

    memcpy(node[N].g,g,sizeof(g));

    node[N].D=D;

}

void solve()

{

    Matrix temp;

    temp.n=temp.m=D;

    memcpy(temp.mat,g,sizeof(g));

    Matrix ans=pow_M(temp,M);

    if(ans.mat[1][0]==0)printf("Impossible\n");

    else printf("%d\n",ans.mat[1][0]%MOD);

}

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    for(int i=0;i<20;i++)node[i].D=0;

    while(scanf("%d%d",&N,&M)==2)

    {

        init();

        solve();

    }

    return 0;

}