Neko and Aki's Prank CodeForces - 1152D (括号序列,dp)

时间：2022-05-13 06:44:32

大意: 将所有长度为2*n的合法括号序列建成一颗trie树, 求trie树上选出一个最大不相交的边集, 输出边集大小.

最大边集数一定不超过奇数层结点数. 这个上界可以通过从底层贪心达到, 所以就转化为求奇数层结点数.

然后就dp求出前$i$为'('比')'多j个的方案数, 奇数层且合法的时候统计一下贡献即可.

#include <iostream>

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#include <bitset>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}

//head

const int N = 4e3+10;

int n, dp[N][N];

int main() {

	scanf("%d", &n);

	dp[0][0] = 1;

	int ans = 0;

	REP(i,0,2*n) {

		REP(j,0,i) {

			(dp[i+1][j+1] += dp[i][j]) %= P;

			if (j) (dp[i+1][j-1] += dp[i][j]) %= P;

			if (2*n-i>=j&&i%2==1) (ans += dp[i][j]) %= P;

		}

	}

	printf("%d\n", ans);

}

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