luogu P5301 [GXOI/GZOI2019]宝牌一大堆
wdnm又是打麻将
首先国土无双可以直接枚举哪种牌用了\(2\)次算贡献,然后\(7\)个对子可以把每种牌的对子贡献排序,取最大的\(7\)个,剩下的牌直接暴力枚举是不行的,考虑dp,设\(f_{i,0\sim1,j,k,0\sim4,0\sim4}\),表示考虑前\(i\)种牌,\(0\sim1\)个对子,\(j\)个\(i-1,i,i+1\)顺子,\(k\)个\(i,i+1,i+2\)顺子,\(0\sim4\)个面子,\(0\sim4\)个杠子,的最大价值,转移枚举下一种牌\(i\)怎么放,可以放杠子,或者放\(i,i+1,i+2\)的顺子同时考虑放对子和刻子,然后最后用对应牌型的dp值更新答案
注意顺子的限制条件,同时建议限制一下状态中牌的总数不超过\(18\)个
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<vector>
#include<cmath>
#include<ctime>
#include<queue>
#include<map>
#include<set>
#define LL long long
#define db double
using namespace std;
const int N=40;
int rd()
{
int x=0,w=1;char ch=0;
while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
return x*w;
}
map<string,int> id;
char cc[3];
void inii()
{
id["B"]=0;
id["1m"]=1;
id["2m"]=2;
id["3m"]=3;
id["4m"]=4;
id["5m"]=5;
id["6m"]=6;
id["7m"]=7;
id["8m"]=8;
id["9m"]=9;
id["1p"]=10;
id["2p"]=11;
id["3p"]=12;
id["4p"]=13;
id["5p"]=14;
id["6p"]=15;
id["7p"]=16;
id["8p"]=17;
id["9p"]=18;
id["1s"]=19;
id["2s"]=20;
id["3s"]=21;
id["4s"]=22;
id["5s"]=23;
id["6s"]=24;
id["7s"]=25;
id["8s"]=26;
id["9s"]=27;
id["E"]=28;
id["S"]=29;
id["W"]=30;
id["N"]=31;
id["Z"]=32;
id["F"]=33;
}
int c[5][5]={{1},{1,1},{1,2,1},{1,3,3,1},{1,4,6,4,1}},cn[N];
bool qb[N];
int gtws[13]={1,9,10,18,19,27,28,29,30,31,32,0,33};
LL f[2][2][3][3][5][5],qdz[N],ans;
int main()
{
inii();
int T=rd();
while(T--)
{
ans=0;
for(int i=0;i<34;++i) cn[i]=4,qb[i]=0;
while(1)
{
scanf("%s",cc);
if(cc[0]=='0') break;
--cn[id[cc]];
}
while(1)
{
scanf("%s",cc);
if(cc[0]=='0') break;
qb[id[cc]]=1;
}
bool fg=1;
for(int i=0;fg&&i<13;++i) fg=cn[gtws[i]];
if(fg)
{
LL sm=13;
for(int i=0;i<13;++i) sm*=(1ll<<qb[gtws[i]])*cn[gtws[i]];
for(int i=0;i<13;++i)
if(cn[gtws[i]]>=2) ans=max(ans,sm*(1ll<<qb[gtws[i]])*(cn[gtws[i]]-1)/2);
}
int m=0;
for(int i=0;i<34;++i)
if(cn[i]>=2) qdz[++m]=(1ll<<(qb[i]*2))*c[cn[i]][2];
if(m>=7)
{
sort(qdz+1,qdz+m+1);
LL sm=7;
for(int i=1;i<=7;++i) sm*=qdz[m-i+1];
ans=max(ans,sm);
}
memset(f,0,sizeof(f));
int nw=1,la=0;
f[la][0][0][0][0][0]=1;
for(int i=0;i<34;++i)
{
for(int j=0;j<=1;++j)
for(int k=0;k<=2;++k)
for(int l=0;l<=2;++l)
for(int p=0;p<=4&&j*2+p*3<=18;++p)
for(int q=0;q<=4&&j*2+p*3+q*4<=18;++q)
{
if(!f[la][j][k][l][p][q]) continue;
int res=cn[i]-k-l,lm=min(i>=1&&i<=27&&(i-1)%9+1<=7?2:0,res);
for(int o=0;o<=lm&&o<=cn[i+1]-l&&o<=cn[i+2]&&p+o<=4&&j*2+(p+o)*3+q*4<=18;++o)
{
LL dt=1ll<<(o*(qb[i]+qb[i+1]+qb[i+2]));
f[nw][j][l][o][p+o][q]=max(f[nw][j][l][o][p+o][q],f[la][j][k][l][p][q]*dt*c[cn[i]][o+k+l]);
if(o+2<=res&&!j&&(j+1)*2+(p+o)*3+q*4<=18) f[nw][j+1][l][o][p+o][q]=max(f[nw][j+1][l][o][p+o][q],f[la][j][k][l][p][q]*dt*c[cn[i]][o+k+l+2]*(1<<(qb[i]*2)));
if(o+3<=res&&p+o+1<=4&&j*2+(p+o+1)*3+q*4<=18) f[nw][j][l][o][p+o+1][q]=max(f[nw][j][l][o][p+o+1][q],f[la][j][k][l][p][q]*dt*c[cn[i]][o+k+l+3]*(1<<(qb[i]*3)));
}
if(res==4&&q+1<=4&&j*2+p*3+(q+1)*4<=18) f[nw][j][l][0][p][q+1]=max(f[nw][j][l][0][p][q+1],f[la][j][k][l][p][q]*c[cn[i]][4]*(1<<(qb[i]*4)));
f[la][j][k][l][p][q]=0;
}
nw^=1,la^=1;
}
ans=max(ans,f[la][1][0][0][4][0]);
ans=max(ans,f[la][1][0][0][3][1]);
ans=max(ans,f[la][1][0][0][2][2]);
ans=max(ans,f[la][1][0][0][1][3]);
ans=max(ans,f[la][1][0][0][0][4]);
printf("%lld\n",ans);
}
return 0;
}
