CF-1152

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D. Neko and Aki’s Prank

题意

将长度为 $2n$ 的所有合法括号匹配放入字典树中，求这棵树的最大的边集，边集里的边两两不相连

题解

对于括号匹配 $(((),()((,(()($ 来说，其子树是一样的， $dp[i][j]$ 表示深度为 $i$ 的左括号比右括号多 $j$ 的括号匹配的答案，我们直接贪心的取边，能取就取，使用一个 $vis[i][j]$ 来标记点是否取了，$dp[i][j]=dp[i+1][j+1]+dp[i+1][j-1]+(vis[i+1][j+1]==0||vis[i+1][j-1]==0)$，注意判断 $[i][j]$ 是否是合法的节点

代码

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#include <bits/stdc++.h>
using namespace std;
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define fro1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define fro0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define memcp(a,b) memcpy(a,b,sizeof(b))
#define oper(type) bool operator <(const type y)const
#define mp make_pair
#define pu_b push_back
#define pu_f push_front
#define po_b pop_back
#define po_f pop_front
#define fi first
#define se second
#define whiel while
#define retrun return
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef double db;
typedef long double ldb;
typedef long long ll;
typedef int itn;
void in(initializer_list<int*> li){for(auto ptr=li.begin();ptr!=li.end();ptr++)scanf("%d",*ptr);}
void in(initializer_list<ll*> li){for(auto ptr=li.begin();ptr!=li.end();ptr++)scanf("%lld",*ptr);}
void in(initializer_list<db*> li){for(auto ptr=li.begin();ptr!=li.end();ptr++)scanf("%lf",*ptr);}
void out(initializer_list<int> li){auto ti=li.end();ti--;for(auto ptr=li.begin();ptr!=li.end();ptr++)printf("%d%c",*ptr,ptr==ti?'\n':' ');}
void out(initializer_list<ll> li){auto ti=li.end();ti--;for(auto ptr=li.begin();ptr!=li.end();ptr++)printf("%lld%c",*ptr,ptr==ti?'\n':' ');}
void out(initializer_list<db> li){auto ti=li.end();ti--;for(auto ptr=li.begin();ptr!=li.end();ptr++)printf("%f%c",*ptr,ptr==ti?'\n':' ');}
void out(int a,bool ln){printf("%d%c",a,ln?'\n':' ');}
void out(ll a,bool ln){printf("%lld%c",a,ln?'\n':' ');}
void out(db a,int digit,bool ln){printf("%.*f%c",digit,a,ln?'\n':' ');}
void out(ldb a,int digit,bool ln){printf("%.*Lf%c",digit,a,ln?'\n':' ');}
void out0(int a[],int n){for0(i,n)out(a[i],i==n-1);}
void out1(int a[],int n){for1(i,n)out(a[i],i==n);}
void out0(ll a[],int n){for0(i,n)out(a[i],i==n-1);}
void out1(ll a[],int n){for1(i,n)out(a[i],i==n);}
int in(int &a,int &b,int &c,int &d){return scanf("%d%d%d%d",&a,&b,&c,&d);}
int in(int &a,int &b,int &c){return scanf("%d%d%d",&a,&b,&c);}
int in(int &a,int &b){return scanf("%d%d",&a,&b);}
int in(ll &a,ll &b,ll &c,ll &d){return scanf("%lld%lld%lld%lld",&a,&b,&c,&d);}
int in(ll &a,ll &b,ll &c){return scanf("%lld%lld%lld",&a,&b,&c);}
int in(ll &a,ll &b){return scanf("%lld%lld",&a,&b);}
int in(ll &a){return scanf("%lld",&a);}
int in(int &a){return scanf("%d",&a);}
int in(char *s){return scanf("%s",s);}
int in(char &c){return scanf("%c",&c);}
int in(db &a){return scanf("%lf",&a);}
int in(ldb &a){return scanf("%Lf",&a);}
void in0(int a[],int n){for0(i,n)in(a[i]);}
void in1(int a[],int n){for1(i,n)in(a[i]);}
void in0(ll a[],int n){for0(i,n)in(a[i]);}
void in1(ll a[],int n){for1(i,n)in(a[i]);}
const db pi = acos(-1);
const db eps = 1e-8;
int sign(db a) {return a < -eps ? -1 : a > eps;}
int db_cmp(db a, db b){ return sign(a-b);}
const ll inf =0x3f3f3f3f;
const ll mod = 1e9+7;
const int M = 2.1e5;
const int N = 2.1e3;
/*-----------------------------------head----------------------------------------------*/

int dp[N][N],n;
bool vis[N][N];
bool check(int i,int j){
    if(j>=0&&j<=n&&(i-j)%2==0&&i+j<=2*n)retrun 1;
    else retrun 0;
}
int main() {
    in(n);
    for(int i=2*n-1;i>=0;i--){
        for(int j=0;j<=min(n,i);j++){
            bool flag=0;
            if(check(i+1,j-1)){
                if(vis[i+1][j-1]==0)vis[i+1][j-1]=1,flag=1;
                dp[i][j]=(dp[i][j]+dp[i+1][j-1])%mod;
            }
            if(check(i+1,j+1)){
                if(vis[i+1][j+1]==0&&flag==0)flag=1,vis[i+1][j+1]=1;
                dp[i][j]=(dp[i][j]+dp[i+1][j+1])%mod;
            }
            if(flag){
                vis[i][j]=1,dp[i][j]=(dp[i][j]+1)%mod;
            }
        }
    }
    out(dp[0][0],1);
    return 0;
}