ZOJ 3256 Tour in the Castle(插头DP-按行递推—矩阵)

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3540

题意：给出一个n*m的方格。从左上角走到左下角每个格子只能走一次。有多少种？



思路：首先，按行递推，得到一个矩阵，表示两个状态是否可以连接。那么最关键的判断两行是否可连接。上一行的一个状态，枚举当前行的插头的状态,就是有无插头，2^n，然后根据上一行的插头的连通性判断当前行插头的连通性。

#include <iostream>
#include <cstdio>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <map>

#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
#define abs(x) ((x)>=0?(x):-(x))
#define i64 long long
#define u32 unsigned int
#define u64 unsigned long long
#define clr(x,y) memset(x,y,sizeof(x))
#define CLR(x) x.clear()
#define ph(x) push(x)
#define pb(x) push_back(x)
#define Len(x) x.length()
#define SZ(x) x.size()
#define PI acos(-1.0)
#define sqr(x) ((x)*(x))

#define FOR0(i,x) for(i=0;i<x;i++)
#define FOR1(i,x) for(i=1;i<=x;i++)
#define FOR(i,a,b) for(i=a;i<=b;i++)
#define FORL0(i,a) for(i=a;i>=0;i--)
#define FORL1(i,a) for(i=a;i>=1;i--)
#define FORL(i,a,b)for(i=a;i>=b;i--)

#define rush() int C; for(scanf("%d",&C);C--;)
#define Rush(n)  while(scanf("%d",&n)!=-1)
using namespace std;

void RD(int &x){scanf("%d",&x);}
void RD(i64 &x){scanf("%lld",&x);}
void RD(u32 &x){scanf("%u",&x);}
void RD(double &x){scanf("%lf",&x);}
void RD(int &x,int &y){scanf("%d%d",&x,&y);}
void RD(u32 &x,u32 &y){scanf("%u%u",&x,&y);}
void RD(double &x,double &y){scanf("%lf%lf",&x,&y);}
void RD(int &x,int &y,int &z){scanf("%d%d%d",&x,&y,&z);}
void RD(u32 &x,u32 &y,u32 &z){scanf("%u%u%u",&x,&y,&z);}
void RD(double &x,double &y,double &z){scanf("%lf%lf%lf",&x,&y,&z);}
void RD(char &x){x=getchar();}
void RD(char *s){scanf("%s",s);}
void RD(string &s){cin>>s;}

void PR(int x) {printf("%d\n",x);}
void PR(i64 x) {printf("%lld\n",x);}
void PR(u32 x) {printf("%u\n",x);}
void PR(double x) {printf("%.4lf\n",x);}
void PR(char x) {printf("%c\n",x);}
void PR(char *x) {printf("%s\n",x);}
void PR(string x) {cout<<x<<endl;}

const int mod=7777777;
const int INF=1000000000;
const int HASHSIZE=100007;
const int N=1000005;

int n,m,code[15],a[130][130],D;

class Matrix
{
public:
int a[130][130];

Matrix operator*(Matrix p)
{
int i,j,k;
i64 temp;
Matrix ans;
FOR0(i,D) FOR0(j,D)
{
temp=0;
FOR0(k,D) temp+=(i64)a[i][k]*p.a[k][j];
ans.a[i][j]=temp%mod;
}
return ans;
}
};

Matrix p,q;

struct node
{
int e,next
,head[HASHSIZE],state
;

void init()
{
clr(head,-1);
e=0;
}

int push(int s)
{
int i,x=s%HASHSIZE;
for(i=head[x];i!=-1;i=next[i])
{
if(state[i]==s) return i;
}
state[e]=s;
next[e]=head[x];
head[x]=e++;
return e-1;
}
};

node dp;

void decode(int code[],int m,int st)
{
int i;
FORL0(i,m-1) code[i]=st&3,st>>=2;
}

int encode(int code[],int m)
{
int ans=0,hash[100],i,cnt=1;
clr(hash,-1);
hash[0]=0;
FOR0(i,m)
{
if(hash[code[i]]==-1) hash[code[i]]=cnt++;
code[i]=hash[code[i]];
ans=(ans<<2)|code[i];
}
return ans;
}

int OK(int st1,int st2)
{
decode(code,n,st1);
int i,j,k,flag=0,cnt=n,temp;
FOR0(i,n)
{
if(!flag)
{
if(!code[i]&&!(st2&(1<<i))) return 0;
if(code[i]&&(st2&(1<<i))) continue;
if(code[i]) flag=code[i];
else flag=-1;
k=i;
}
else
{
if(code[i]&&(st2&(1<<i)))return 0;
if(!code[i]&&!(st2&(1<<i))) continue;
if(code[i])
{
if(code[i]==flag)
{
if(!(st2==0&&i==n-1)) return 0;
}
if(flag>0)
{
temp=code[i];
FOR0(j,n) if(code[j]==temp) 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]=cnt++;
}
flag=0;
}
}
if(flag) return 0;
return 1;
}

void init()
{
dp.init(); dp.push(0); clr(code,0);
code[0]=code[n-1]=1; dp.push(encode(code,n));
clr(a,0);
int i,j,k;
FOR1(i,dp.e-1)
{
FOR0(j,(1<<n)) if(OK(dp.state[i],j))
{
k=dp.push(encode(code,n));
a[i][k]=1;
}
}
D=dp.e;
}

int main()
{
Rush(n)
{
RD(m); init(); clr(p.a,0);
int i,j;
FOR0(i,D) p.a[i][i]=1;
FOR0(i,D) FOR0(j,D) q.a[i][j]=a[i][j];
while(m)
{
if(m&1) p=p*q;
q=q*q;
m>>=1;
}
if(p.a[1][0]) PR(p.a[1][0]);
else puts("Impossible");
}
return 0;
}