CF14Eラクダ

時間の複雑さ：O（64n t ^ 2）

#include <bits/stdc++.h>
#define int long long
using namespace std;
int n,t,ans;
int f[21][5][5][12][12];
signed main(){
    
    
	//f[i][j][k][t1][t2]:前i位，第i位为j，第i-1位为k，前i-1位，t1个条件1，t2个条件2已经满足 
	scanf("%lld%lld",&n,&t);
	for (register int i=1; i<=4; ++i)
	for (register int j=1; j<=4; ++j) if (i!=j) f[2][i][j][0][0]=1; 
	for (register int l=3; l<=n; ++l)	
	for (register int t1=0; t1<=t; ++t1)
	for (register int t2=0; t2<t; ++t2)	
	for (register int i=1; i<=4; ++i)
	for (register int j=1; j<=4; ++j)
	for (register int k=1; k<=4; ++k)
	{
    
    
		if (k<j && j>i && t1+1<=t) f[l][i][j][t1+1][t2]+=f[l-1][j][k][t1][t2];
		else if (k>j && j<i && t2+1<=t) f[l][i][j][t1][t2+1]+=f[l-1][j][k][t1][t2];
		else if (k!=j && j!=i && k!=i) f[l][i][j][t1][t2]+=f[l-1][j][k][t1][t2];
	}
	for (register int i=1; i<=4; ++i)
	for (register int j=1; j<=4; ++j) if (i!=j) ans+=f[n][i][j][t][t-1];
	printf("%lld\n",ans);
return 0;	
}