# circuits

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 418    Accepted Submission(s): 141

Problem Description
Given a map of N * M (2 <= N, M <= 12) , '.' means empty, '*' means walls. You need to build K circuits and no circuits could be nested in another. A circuit is a route connecting adjacent cells in a cell sequence, and also connect the first cell and the last cell. Each cell should be exactly in one circuit. How many ways do we have?

Input
The first line of input has an integer T, number of cases.
For each case:
The first line has three integers N M K, as described above.
Then the following N lines each has M characters, ‘.’ or ‘*’.

Output
For each case output one lines.
Each line is the answer % 1000000007 to the case.

Sample Input
2 4 4 1 **.. .... .... .... 4 4 1 .... .... .... ....

Sample Output
2 6

Source

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/*
HDU 4285

G++ 11265ms  11820K
C++ 10656ms  11764K

*/

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

const int MAXD=15;
const int STATE=1000010;
const int HASH=300007;//这个大一点可以防止TLE,但是容易MLE
const int MOD=1000000007;

int N,M,K;
int maze[MAXD][MAXD];
int code[MAXD];
int ch[MAXD];
int num;//圈的个数
struct HASHMAP
{
long long state[STATE];
int f[STATE];
void init()
{
size=0;
}
void push(long long st,int ans)
{
int i;
int h=st%HASH;
if(state[i]==st)
{
f[i]+=ans;
f[i]%=MOD;
return;
}
state[size]=st;
f[size]=ans;
}
}hm[2];
void decode(int *code,int m,long long  st)
{
num=st&63;
st>>=6;
for(int i=m;i>=0;i--)
{
code[i]=st&7;
st>>=3;
}
}
long long encode(int *code,int m)//最小表示法
{
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];
}
st<<=6;
st|=num;
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(num>=K)continue;
int t=0;
//要避免环套环的情况，需要两边插头数为偶数
for(int p=0;p<j-1;p++)
if(code[p])t++;
if(t&1)continue;
if(num<K)
{
num++;
code[j-1]=code[j]=0;
hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k]);
}
}
else
{
code[j-1]=code[j]=0;
for(int t=0;t<=M;t++)
if(code[t]==up)
code[t]=left;
hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k]);
}
}
else if(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]=0;
code[j-1]=t;
hm[cur^1].push(encode(code,j==M?M-1: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,j==M?M-1: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;
hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k]);
}
}
char str[20];
void init()
{
scanf("%d%d%d",&N,&M,&K);
memset(maze,0,sizeof(maze));
for(int i=1;i<=N;i++)
{
scanf("%s",&str);
for(int j=1;j<=M;j++)
if(str[j-1]=='.')
maze[i][j]=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;
}
int ans=0;
for(i=0;i<hm[cur].size;i++)
if(hm[cur].state[i]==K)
{
ans+=hm[cur].f[i];
ans%=MOD;
}
printf("%d\n",ans);

}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
scanf("%d",&T);
while(T--)
{
init();
solve();
}
return 0;
}

/*
Sample Input
2
4 4 1
**..
....
....
....
4 4 1
....
....
....
....

Sample Output
2
6

*/

/*
HDU  4285

G++ 11765ms  12560K
C++  11656ms 12504K
*/

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

const int MAXD=15;
const int STATE=1000010;
const int HASH=100007;
const int MOD=1000000007;

int N,M,K;
int maze[MAXD][MAXD];
int code[MAXD];
int ch[MAXD];

struct HASHMAP
{
long long state[STATE];
int f[STATE];
int cir[STATE];//形成的圈的个数
void init()
{
size=0;
}
void push(long long st,int ans,int _cir)
{
int i,h=st%HASH;
if(state[i]==st&&cir[i]==_cir)
{
f[i]+=ans;
f[i]%=MOD;
return;
}
state[size]=st;
f[size]=ans;
cir[size]=_cir;
}
}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)//最小表示法
{
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(hm[cur].cir[k]>=K)continue;
int t=0;
for(int p=0;p<j-1;p++)
if(code[p])t++;
if(t&1)continue;
if(hm[cur].cir[k]<K)
{
code[j-1]=code[j]=0;
hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k],hm[cur].cir[k]+1);
}
}
else
{
code[j-1]=code[j]=0;
for(int t=0;t<=M;t++)
if(code[t]==up)
code[t]=left;
hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k],hm[cur].cir[k]);
}
}
else if(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],hm[cur].cir[k]);
}
if(maze[i+1][j])
{
code[j]=0;
code[j-1]=t;
hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k],hm[cur].cir[k]);
}
}
else
{
if(maze[i][j+1]&&maze[i+1][j])
{
code[j-1]=code[j]=13;
hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k],hm[cur].cir[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;
hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].f[k],hm[cur].cir[k]);
}
}
char str[20];
void init()
{
scanf("%d%d%d",&N,&M,&K);
memset(maze,0,sizeof(maze));
for(int i=1;i<=N;i++)
{
scanf("%s",&str);
for(int j=1;j<=M;j++)
if(str[j-1]=='.')
maze[i][j]=1;
}
}
void solve()
{
int i,j,cur=0;
hm[cur].init();
hm[cur].push(0,1,0);
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;
}
int ans=0;
for(i=0;i<hm[cur].size;i++)
if(hm[cur].cir[i]==K)
{
ans+=hm[cur].f[i];
ans%=MOD;
}
printf("%d\n",ans);

}
int main()
{
//  freopen("in.txt","r",stdin);
//   freopen("out.txt","w",stdout);
int T;
scanf("%d",&T);
while(T--)
{
init();
solve();
}
return 0;
}

/*
Sample Input
2
4 4 1
**..
....
....
....
4 4 1
....
....
....
....

Sample Output
2
6

*/

posted on 2012-10-02 13:51  kuangbin  阅读(1262)  评论(2编辑  收藏  举报

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