BZOJ1814: Ural 1519 Formula 1(插头Dp)
Description
Regardless of the fact, that Vologda could not get rights to hold the Winter Olympic games of 20**, it is well-known, that the city will conduct one of the Formula 1 events. Surely, for such an important thing a new race circuit should be built as well as hotels, restaurants, international airport - everything for Formula 1 fans, who will flood the city soon. But when all the hotels and a half of the restaurants were built, it appeared, that at the site for the future circuit a lot of gophers lived in their holes. Since we like animals very much, ecologists will never allow to build the race circuit over the holes. So now the mayor is sitting sadly in his office and looking at the map of the circuit with all the holes plotted on it. Problem Who will be smart enough to draw a plan of the circuit and keep the city from inevitable disgrace? Of course, only true professionals - battle-hardened programmers from the first team of local technical university!.. But our heroes were not looking for easy life and set much more difficult problem: "Certainly, our mayor will be glad, if we find how many ways of building the circuit are there!" - they said. It should be said, that the circuit in Vologda is going to be rather simple. It will be a rectangle N*M cells in size with a single circuit segment built through each cell. Each segment should be parallel to one of rectangle's sides, so only right-angled bends may be on the circuit. At the picture below two samples are given for N = M = 4 (gray squares mean gopher holes, and the bold black line means the race circuit). There are no other ways to build the circuit here. 一个 m * n 的棋盘,有的格子存在障碍,求经过所有非障碍格子的哈密顿回路个数
Input
The first line contains the integer numbers N and M (2 ≤ N, M ≤ 12). Each of the next N lines contains M characters, which are the corresponding cells of the rectangle. Character "." (full stop) means a cell, where a segment of the race circuit should be built, and character "*" (asterisk) - a cell, where a gopher hole is located.
Output
You should output the desired number of ways. It is guaranteed, that it does not exceed 2^63-1.
Sample Input
4 4
**..
....
....
....
**..
....
....
....
Sample Output
2
解题思路：
发现一个格点一定有一进一出，就可以3进制表示括号序列轮廓线Dp就好了。
把该存的存入map/Hash表中，元素访问状态转移，要写好编码和解码。3进制改用4进制。
分类讨论一下，10种情况，画个图就好了。
方程只关心最后一个格点，可以滚动掉2维。
代码：(Hash挂链)
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
typedef long long lnt;
typedef unsigned int uit;
typedef long long unt;
struct int_2{
uit x;
unt y;
int_2(){}
int_2(uit a,unt b){x=a;y=b;}
};
class Fast_map{
#define mod 2333
public:
bool find(uit x)
{
int y=x%mod;
for(int i=;i<to[y].size();i++)
if(to[y][i].x==x)
return true;
return false;
}
void ins(uit x,unt v)
{
int f=x%mod;
to[f].push_back(int_2(x,v));
return ;
}
void add(uit x,unt v)
{
int f=x%mod;
for(int i=;i<to[f].size();i++)
if(to[f][i].x==x)
{
to[f][i].y+=v;
return ;
}
return ;
}
void clear(void)
{
for(int i=;i<mod;i++)
to[i].clear();
return ;
}
void start(void)
{
jjj=;
iii=;
while(iii<mod&&!to[iii].size())
iii++;
return ;
}
bool going_on(void)
{ return iii<mod;
}
void go(void)
{
jjj++;
if(jjj<to[iii].size())
return ;
jjj=;
iii++;
while(iii<mod&&!to[iii].size())
iii++;
return ;
}
int_2 ite(void)
{
return to[iii][jjj];
}
void move(void)
{
for(int i=;i<mod;i++)
for(int j=;j<to[i].size();j++)
to[i][j].x<<=;
return ;
}
private:
std::vector<int_2>to[mod];
int iii,jjj;
#undef mod
}dp[];
int n,m;
int p,q;
int ei,ej;
unt ans;
char cmd[];
bool mp[][];
void update(int i,uit j,unt val)
{
if(!dp[i].find(j))
dp[i].ins(j,val);
else
dp[i].add(j,val);
return ;
}
uit change(uit x,int plc,uit val)
{
uit ans=x;
uit cut=0ul-1ul;
cut^=(<<(plc<<));
cut^=(<<(plc<<|));
ans&=cut;
ans|=(val<<(plc<<));
return ans;
}
int main()
{
memset(mp,,sizeof(mp));
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++)
{
scanf("%s",cmd+);
for(int j=;j<=m;j++)
{
if(cmd[j]=='.')
{
mp[i][j]=true;
ei=i,ej=j;
}
}
}
p=,q=;
dp[].ins(,);
for(int i=;i<=n;i++)
{
for(int j=;j<=m;j++)
{
std::swap(p,q);
dp[p].clear();
for(dp[q].start();dp[q].going_on();dp[q].go())
{
int_2 state=dp[q].ite();
uit s=state.x;
unt v=state.y;
uit tl=(s>>((j-)<<))&;
uit tu=(s>>(j<<))&;
if(!mp[i][j])
{
if(tl|tu);else
update(p,s,v);
}else{
uit ts;
if(tl==)
{
if(tu==)
{
if(mp[i][j+]&&mp[i+][j]);else continue;
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}else if(tu==)
{
if(mp[i+][j])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
if(mp[i][j+])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
}else{
if(mp[i+][j])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
if(mp[i][j+])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
}
}else if(tl==)
{
if(tu==)
{
if(mp[i+][j])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
if(mp[i][j+])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
}else if(tu==)
{
int cnt=;
for(int kt=j+;kt<=m;kt++)
{
if(((s>>(kt<<))&)==)
cnt++;
if(((s>>(kt<<))&)==)
cnt--;
if(!cnt)
{
ts=change(s,kt,);
break;
}
}
ts=change(ts,j-,);
ts=change(ts,j,);
update(p,ts,v);
}else{
if(i==ei&&j==ej)
ans+=v;
}
}else{
if(tu==)
{
if(mp[i+][j])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
if(mp[i][j+])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
}else if(tu==)
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}else{
int cnt=-;
for(int kt=j-;kt>=;kt--)
{
if(((s>>(kt<<))&)==)
cnt++;
if(((s>>(kt<<))&)==)
cnt--;
if(!cnt)
{
ts=change(s,kt,);
break;
}
}
ts=change(ts,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
}
}
}
}
dp[p].move();
}
printf("%lld\n",ans);
return ;
}
代码(map)
#include<map>
#include<cstdio>
#include<cstring>
#include<algorithm>
typedef long long lnt;
typedef unsigned int uit;
typedef long long unt;
using std::make_pair;
int n,m;
int p,q;
int ei,ej;
unt ans;
char cmd[];
bool mp[][];
std::map<uit,unt>dp[],ooo;
void update(int i,uit j,unt val)
{
if(dp[i].find(j)==dp[i].end())
dp[i][j]=val;
else
dp[i][j]+=val;
return ;
}
uit change(uit x,int plc,uit val)
{
uit ans=x;
uit cut=0ul-1ul;
cut^=(<<(plc<<));
cut^=(<<(plc<<|));
ans&=cut;
ans|=(val<<(plc<<));
return ans;
}
int main()
{
memset(mp,,sizeof(dp));
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++)
{
scanf("%s",cmd+);
for(int j=;j<=m;j++)
{
if(cmd[j]=='.')
{
mp[i][j]=true;
ei=i,ej=j;
}
}
}
p=,q=;
dp[][]=;
for(int i=;i<=n;i++)
{
for(int j=;j<=m;j++)
{
std::swap(p,q);
dp[p].clear();
for(std::map<uit,unt>::iterator State=dp[q].begin();State!=dp[q].end();State++)
{
uit s=State->first;
unt v=State->second;
uit tl=(s>>((j-)<<))&;
uit tu=(s>>(j<<))&;
if(!mp[i][j])
{
if(tl|tu);else
update(p,s,v);
}else{
uit ts;
if(tl==)
{
if(tu==)
{
if(mp[i][j+]&&mp[i+][j]);else continue;
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}else if(tu==)
{
if(mp[i+][j])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
if(mp[i][j+])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
}else{
if(mp[i+][j])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
if(mp[i][j+])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
}
}else if(tl==)
{
if(tu==)
{
if(mp[i+][j])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
if(mp[i][j+])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
}else if(tu==)
{
int cnt=;
for(int kt=j+;kt<=m;kt++)
{
if(((s>>(kt<<))&)==)
cnt++;
if(((s>>(kt<<))&)==)
cnt--;
if(!cnt)
{
ts=change(s,kt,);
break;
}
}
ts=change(ts,j-,);
ts=change(ts,j,);
update(p,ts,v);
}else{
if(i==ei&&j==ej)
ans+=v;
}
}else{
if(tu==)
{
if(mp[i+][j])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
if(mp[i][j+])
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
}else if(tu==)
{
ts=change(s,j-,);
ts=change(ts,j,);
update(p,ts,v);
}else{
int cnt=-;
for(int kt=j-;kt>=;kt--)
{
if(((s>>(kt<<))&)==)
cnt++;
if(((s>>(kt<<))&)==)
cnt--;
if(!cnt)
{
ts=change(s,kt,);
break;
}
}
ts=change(ts,j-,);
ts=change(ts,j,);
update(p,ts,v);
}
}
}
}
}
ooo.clear();
for(std::map<uit,unt>::iterator j=dp[p].begin();j!=dp[p].end();j++)
if(!((j->first)>>(m<<)))
ooo[((j->first)<<)]=j->second;
dp[p].clear();
for(std::map<uit,unt>::iterator j=ooo.begin();j!=ooo.end();j++)
dp[p][j->first]=j->second;
}
printf("%lld\n",ans);
return ;
}
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