topic
Thinking
This question is the main subject of the bad state of want and not well understood
If you and I had no contact with the same Mahjong
We first performed the entire mahjong hash processing
Defined \ (dp_ {i, j, k, l, m, n} \) front the i cards, the number of taping and face a total of j, the number of birds have to k, of \ (i-2 \) species number of cards is l, of \ (i-1 \) number of card types is m, the number of kinds of cards is n i
To note that the value of the DP is not i-2 ~ i into consideration the three cards
You can write transfer
\ (\ begin {cases} dp_ {i, j + 1, k, l + 1, m + 1, n + 1} (\ mbox { straight}) \\ dp_ {i, j + 1, k, l , m, n + 3} (pung) \\ dp_ {i, j + 1, k, l, m, n + 4} ( first bar) \\ dp_ {i, j, k + 1, l, m , n + 2} (The birds) \ Cases} End {\) = \ (dp_ {I, J, K, L, m, n-} \)
Direct transfer place, because we \ (dp \) is not considered i-2 ~ i kind of chess
In another case
\(dp_{i+1,j,k,m,n,0}=dp_{i,j,k,m,n,l}*C_{t_{i}}^l*qkpow(bp_i,l)\)
Wherein \ (t_i \) represents the remaining number of the i-chess, \ (bp_i = \ Cases the begin {2} [Po is the i-th number of cards] \\ 1 [i-th-card is not] \ end {cases} \)
When the significance of the equation would not consider the i chess, the current \ (dp \) state back transfer, is also very good understanding of
Do not forget to add the last contribution statistics about statistics when three last chess
Then talk about optimization
The current state is 0 can skip
Proof: The last state must be a product of formula, wherein when a is 0, 0 is directly entire
Can not be considered thick stick
Proof: \ (^ 3 = 4 C_4 and C_4 ^ 4 = 1 \) means that even if this card is a treasure card is not worthwhile
But this does not mean that the condition will become cycle, we just continue more than a few state
In fact constant level decreases
But you can make 50 points helicopter from 100 points
This question is really good road (cancer) title
The importance of constant glance
Code
#pragma GCC optimize(2)
#include<iostream>
#include<cstring>
#include<climits>
#include<queue>
using namespace std;
int T;
priority_queue<int> q;
long long ans;
long long tmp;
int num_pow[3][5];
long long dp[35][5][2][5][5][5];
int sz[35]={0,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,0,0,0,0,0,0,0};
int gs[14]={0,1,9,10,18,19,27,28,29,30,31,32,33,34};
int t[35];
int bp[35];
int c[5][5]=
{
{1,0,0,0,0},
{1,1,0,0,0},
{1,2,1,0,0},
{1,3,3,1,0},
{1,4,6,4,1}
};
string a;
int solve_hash(string a)
{
if(a[1]=='m')
return a[0]-'0';
if(a[1]=='p')
return a[0]-'0'+9;
if(a[1]=='s')
return a[0]-'0'+18;
if(a[0]=='E')
return 28;
if(a[0]=='S')
return 29;
if(a[0]=='W')
return 30;
if(a[0]=='N')
return 31;
if(a[0]=='Z')
return 32;
if(a[0]=='B')
return 33;
if(a[0]=='F')
return 34;
}
void init()
{
while(!q.empty())
q.pop();
memset(dp,0,sizeof(dp));
dp[1][0][0][0][0][0]=1;
ans=0;
for(int i=1;i<=34;i++)
{
bp[i]=1;
t[i]=4;
}
}
int qkpow(int a,int b)
{
if(num_pow[a][b])
return num_pow[a][b];
if(b==0)
return 1;
if(b==1)
return a;
long long t=qkpow(a,b/2);
t=(t*t);
if(b%2==1)
t*=a;
num_pow[a][b]=t;
return t;
}
void c_in()
{
init();
while(1)
{
cin>>a;
if(a[0]=='0')
break;
t[solve_hash(a)]--;
}
while(1)
{
cin>>a;
if(a[0]=='0')
break;
bp[solve_hash(a)]=2;
}
for(int i=1;i<=13;i++)
{
tmp=1;
for(int j=1;j<=13;j++)
{
if(i==j)
{
if(t[gs[i]]<2)
tmp=0;
else
tmp=tmp*c[t[gs[i]]][2]*qkpow(bp[gs[i]],2);
}
else
{
if(!t[gs[j]])
tmp=0;
else
tmp=tmp*c[t[gs[j]]][1]*qkpow(bp[gs[j]],1);
}
}
ans=max(ans,tmp*13);
}
for(int i=1;i<=34;i++)
if(t[i]>=2)
q.push(c[t[i]][2]*qkpow(bp[i],2));
if(q.size()>=7)
{
tmp=1;
for(int i=1;i<=7;i++)
{
tmp=tmp*q.top();
q.pop();
}
ans=max(ans,tmp*7);
}
for(int i=1;i<=34;i++)
{
for(int j=0;j<=4;j++)
{
for(int k=0;k<=1;k++)
{
for(int l=0;l<=4;l++)
{
for(int m=0;m<=4;m++)
{
for(int n=0;n<=4;n++)
{
long long now=dp[i][j][k][l][m][n];
if(!now)
continue;
if(k==0&&t[i]-n>=2)
dp[i][j][k+1][l][m][n+2]=max(dp[i][j][k+1][l][m][n+2],now);
if(t[i]-n>=3&&j<4)
dp[i][j+1][k][l][m][n+3]=max(dp[i][j+1][k][l][m][n+3],now);
if(sz[i]&&t[i-2]-l>=1&&t[i-1]-m>=1&&t[i]-n>=1&&j<4)
dp[i][j+1][k][l+1][m+1][n+1]=max(dp[i][j+1][k][l+1][m+1][n+1],now);
if(i<34)
dp[i+1][j][k][m][n][0]=max(dp[i+1][j][k][m][n][0],now*(i>2?c[t[i-2]][l]:1)*qkpow(i>2?bp[i-2]:1,l));
if(i==34&&j==4&&k==1)
ans=max(ans,now*c[t[i-2]][l]*qkpow(bp[i-2],l)*c[t[i-1]][m]*qkpow(bp[i-1],m)*c[t[i]][n]*qkpow(bp[i],n));
}
}
}
}
}
}
cout<<ans<<'\n';
}
int main()
{
ios::sync_with_stdio(false);
cin>>T;
for(int i=1;i<=T;i++)
c_in();
return 0;
}Yes that is six-dimensional