HDU 4285 circuits（插头DP）
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4285
题意：给出一个n*m的方格，有.和*两种符号。用所有的.围成K个回路，回路不相互包含，有多少种？
思路：至于怎么判断不相互包含，网上说每次左右插头相遇时，判断左插头左侧是不是有偶数个插头，是偶数就能合并这两个插头.
#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 i64 mod=1000000007;
const int HASHSIZE=100007;
const int N=1000005;
int n,m,code[15],cur,pre,sx,sy;
char s[15][15];
i64 ans;
struct node
{
int e,next[N],head[HASHSIZE];
i64 cnt[N],state[N];
void init()
{
clr(head,-1);
e=0;
}
void push(i64 s,i64 val)
{
int i,x=s%HASHSIZE;
for(i=head[x];i!=-1;i=next[i])
{
if(state[i]==s)
{
cnt[i]+=val;
cnt[i]%=mod;
return;
}
}
state[e]=s;
cnt[e]=val;
next[e]=head[x];
head[x]=e++;
}
};
node dp[2];
int num,K;
void decode(int code[],int m,i64 st)
{
num=st&63; st>>=6;
int i;
FORL0(i,m) code[i]=st&7,st>>=3;
}
i64 encode(int code[],int m)
{
i64 ans=0;
int hash[15],i,cnt=1;
clr(hash,-1);
hash[0]=0;
FOR0(i,m+1)
{
if(hash[code[i]]==-1) hash[code[i]]=cnt++;
code[i]=hash[code[i]];
ans=(ans<<3)|code[i];
}
ans=(ans<<6)|num;
return ans;
}
void update1(int i,int j)
{
int x,y,k,t,p,q=j==m?m-1:m;
i64 c;
FOR0(k,dp[pre].e)
{
decode(code,m,dp[pre].state[k]);
c=dp[pre].cnt[k];
x=code[j-1];
y=code[j];
if(x&&x==y)
{
if(num>=K) continue;
p=0;
FOR0(t,j-1) if(code[t]) p++;
if(p&1) continue;
code[j-1]=code[j]=0;
num++;
dp[cur].push(encode(code,q),c);
}
else if(x&&y&&x!=y)
{
code[j-1]=code[j]=0;
FOR0(t,m+1) if(code[t]==y) code[t]=x;
dp[cur].push(encode(code,q),c);
}
else if(x||y)
{
if(i<n&&s[i+1][j])
{
code[j-1]=x+y;
code[j]=0;
dp[cur].push(encode(code,q),c);
}
if(j<m&&s[i][j+1])
{
code[j-1]=0;
code[j]=x+y;
dp[cur].push(encode(code,q),c);
}
}
else if(x==0&&y==0)
{
if(num>=K) continue;
if(i<n&&j<m&&s[i][j+1]&&s[i+1][j])
{
code[j-1]=code[j]=13;
dp[cur].push(encode(code,q),c);
}
}
}
}
void update2(int i,int j)
{
int k,q=j==m?m-1:m;
FOR0(k,dp[pre].e)
{
decode(code,m,dp[pre].state[k]);
if(code[j-1]||code[j]) continue;
dp[cur].push(encode(code,q),dp[pre].cnt[k]);
}
}
i64 DP()
{
ans=0; pre=0,cur=1;
dp[0].init(); dp[0].push(0,1);
int i,j;
FOR1(i,n) FOR1(j,m)
{
dp[cur].init();
if(s[i][j]) update1(i,j);
else update2(i,j);
pre^=1;
cur^=1;
}
FOR0(i,dp[pre].e) if(dp[pre].state[i]==K)
{
ans+=dp[pre].cnt[i];
}
return ans%mod;
}
int main()
{
rush()
{
RD(n,m,K);
int i,j;
FOR1(i,n) RD(s[i]+1);
FOR1(i,n) FOR1(j,m) s[i][j]=(s[i][j]=='.');
FOR1(i,n) FOR1(j,m) if(s[i][j])
{
sx=i,sy=j;
}
PR(DP());
}
return 0;
}