Codeforces Round #224 (Div. 2) E.Ksenia and Combinatorics

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树形DP

dp[ i ][ j ][ g ]表示总共有 i 个节点，匹配数为 j 的时候，有没有把根节点匹配进去的方案数

dp[ i ][ j ][ 0 ] += rela * (  dp[ q ][ k1 ][ 1 ] * dp[ i -1-q ][ k2 ][ 1 ])

dp[ i ][ j ][ 1 ] += (dp[ q ][ k1 ][ 0 ] * dp[ i-1-q ][ k2 ][ 1 ]                        + dp[ q ][ k1 ][ 1 ] * dp[ i-1-q ][ k2 ][ 0 ]+ dp[ q ][ k1 ][ 0 ] * dp[ i-1-q ][ k2 ][ 0 ]) *rela

ｑ表示左子树节点数，i-q-1表示右子树节点数，k1表示左边匹配，k2表示右边匹配

rela表示左子树和右子树节点数的组合

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</pre></p><pre name="code" class="cpp">#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#define CLR(x,y) memset(x,y,sizeof(x))#define eps 1e-9#define INF 0x3f3f3f3fusing namespace std;typedef long long ll;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;template<class T>inline bool read(T &n){    T x = 0, tmp = 1;    char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------const int MOD=1000000007;const int MAXN=60;ll dp[MAXN][MAXN][2],C[105][105];;int n,k;int main(){    while(read(n)&&read(k))    {        CLR(dp,0);        dp[1][0][0]=1;        dp[1][0][1]=0;        dp[0][0][0]=0;        dp[0][0][1]=1;        C[0][0]=1;        for(int i=1; i<=n; i++)        {            for(int j=0; j<=i; j++)            {                if(!j || j==i) C[i][j]=1LL;                else C[i][j]=(C[i-1][j]+C[i-1][j-1])%MOD;            }        }        for(int i=2; i<=n; i++)        {            for(int j=0; j<=k; j++)            {                dp[i][j][0]=(dp[i][j][0]+((i-1)*dp[i-1][j][1])%MOD)%MOD;                if(j)                    dp[i][j][1]=(dp[i][j][1]+((i-1)*dp[i-1][j-1][0])%MOD)%MOD;                for(int q=1; q<i-1; q++)                {                    ll rela=(C[i-1][2]*C[i-3][q-1])%MOD;                    for(int k1=0; k1<=j; k1++)                    {                        int k2=j-k1;                        dp[i][j][0]=(dp[i][j][0]+rela*((dp[q][k1][1]*dp[i-1-q][k2][1])%MOD)%MOD)%MOD;                        if(--k2<0)  continue;                        dp[i][j][1]=(dp[i][j][1]+(dp[q][k1][0]*dp[i-1-q][k2][1] %MOD                                                + dp[q][k1][1]*dp[i-1-q][k2][0] %MOD                                                + dp[q][k1][0]*dp[i-1-q][k2][0])%MOD*rela%MOD)%MOD;                    }                }            }        }        ll ans=(dp[n][k][0]+dp[n][k][1])%MOD;        printf("%lld\n",ans);    }    return 0;}

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