因为题目规定
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任意相邻的两步,玩家不同(一黑一白);
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任意相邻的两步,不能下在同一个劫争处。 我们只需要对每个操作枚举一下 口胡一下:对于第个人,下在第个位置,这一轮他的最有情况一定是另外一个人,下在除了以外的其他位置的最优情况下+1
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
# include<iostream>
# include<iomanip>
# include<algorithm>
# include<cmath>
# include<cstdio>
# include<set>
# include<stack>
# include<queue>
# include<map>
# include<string>
# include<cstring>
# define eps 1e-9
# define fi first
# define se second
# define ll long long
# define int ll
// cout<<fixed<<setprecision(n)
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int > PII;
const int mod=1e9+7;
const int MAX=1e5+10;
const int Time=86400;
const int X=131;
const int inf=0x3f3f3f3f;
const double PI = 1e-4;
double pai = 3.14159265358979323846;
int T,n,dp[MAX],pre[2][15];
void solve(){
int ans = 0;
cin >> n;
memset(dp,0,sizeof dp);
memset(pre,0,sizeof pre);
for(int i = 1 ; i <= n ; i ++ ){
int c,a;
cin >> c >> a;
for(int j = 0 ; j < 2 ; j ++ )
if(j != c){
for(int k = 1 ; k <= 10 ; k ++ )
if(k != a) dp[i] = max(dp[i],dp[pre[j][k]] + 1);
}
pre[c][a] = i;
ans = max(ans,dp[i]);
}
cout<<ans<<"\n";
}
signed main(){
std::ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> T;
while(T--){
solve();
}
return 0;
}