插头DP

丢个代码，就跑。。。暂时弃更。。。

#include<cstdio>
#include<iostream>
#include<cstring>
#define MAXN 15
#define int long long
using namespace std;
int N,M;
bool za[MAXN][MAXN];
char rn[MAXN];
int bin[15];
int ans=0,endx,endy;
const int mo=299987;
struct hhash{
	int st,nt,val;
}bl[2][14348907+10];int hd[2][mo+10],itot[2];
//14348907 shi 3^15
void insert(int st,int num,int knd){
	int u=st%mo+1;
	bool ok=0;
	for(int i=hd[knd][u];i;i=bl[knd][i].nt)
		if(bl[knd][i].st==st){
			bl[knd][i].val+=num;
			ok=1;
			break;
		}
	if(!ok){
		bl[knd][++itot[knd]].st=st;
		bl[knd][itot[knd]].nt=hd[knd][u];
		bl[knd][itot[knd]].val=num;
		hd[knd][u]=itot[knd];
	}
}
signed main(){
//	freopen("da.in","r",stdin);
	bin[0]=1;
	for(int i=1;i<=14;++i)
		bin[i]=bin[i-1]<<2;
	scanf("%lld%lld",&N,&M);
	for(int i=1;i<=N;++i){
		scanf("%s",rn+1);
		for(int j=1;j<=M;++j)
			if(rn[j]=='.'){
				za[i][j]=1;
				endx=i,endy=j;
			}
	}
	int now=1,past=0;
	insert(0,1,past);
	for(int i=1;i<=N;++i){
		for(int s=1;s<=itot[past];++s)
			bl[past][s].st<<=2;
		for(int j=1;j<=M;++j){
			//cout<<"QwQ"<<i<<" "<<j<<endl;
			memset(hd[now],0,sizeof(hd[now]));
			itot[now]=0;
			for(int s=1;s<=itot[past];++s){
				int st=bl[past][s].st,num=bl[past][s].val;
				int b1=(st>>(2*(j-1)))%4,b2=(st>>(2*j))%4;
				//cout<<"orz"<<b1<<" "<<b2<<endl;
				if(!za[i][j]){
					if(!b1&&!b2){
						insert(st,num,now);
					}
				}
				else if(!b1&&!b2){
					if(za[i+1][j]&&za[i][j+1])
						insert(st+bin[j-1]+bin[j]*2,num,now);
				}
				else if(!b1&&b2){
					if(za[i][j+1])
						insert(st,num,now);
					if(za[i+1][j])
						insert(st+b2*bin[j-1]-b2*bin[j],num,now);
				}
				else if(b1&&!b2){
					if(za[i][j+1])
						insert(st-b1*bin[j-1]+b1*bin[j],num,now);
					if(za[i+1][j])
						insert(st,num,now);
				}
				else if(b1==1&&b2==1){
					int dx=1;
					for(int k=j+1;k<=M;++k){
						if((st>>(k*2))%4==1)
							++dx;
						if((st>>(k*2))%4==2)
							--dx;
						if(!dx){
							insert(st-bin[j-1]-bin[j]-bin[k],num,now);
							break;
						}	
					}
				}
				else if(b1==2&&b2==2){
					int dx=1;
					for(int k=j-2;k>=0;--k){
						if((st>>(k*2))%4==2)
							++dx;
						if((st>>(k*2))%4==1)
							--dx;
						if(!dx){
							insert(st-2*bin[j-1]-2*bin[j]+bin[k],num,now);
							break;
						}
					}
				}
				else if(b1==2&&b2==1){
					insert(st-2*bin[j-1]-bin[j],num,now);
				}
				else if(b1==1&&b2==2){
					if(i==endx&&j==endy)
						ans+=num;
				}
			}
			now=past;past=now^1;
		}
	}
	printf("%lld\n",ans);
}
/*
	0 wu cha tou
	1 xia cha tou 
	2 you cha tou
*/

To be continued...

标签：插头,bin,int,knd,st,bl,num,DP
来源： https://www.cnblogs.com/2018hzoicyf/p/11260798.html