[树形DP] BZOJ 4824 [Cqoi2017]老C的键盘

来源：互联网发布：二级vb软件序列号编辑：程序博客网时间：2024/06/04 17:40

首先这是棵完全二叉树
直接fu,i表示u在子树u中排名i的方案数
然后合并就是两个排列合并咯组合数乘一下大于小于都差不多

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}inline void read(char *s){  char c=nc(); int len=0;  for (;!(c=='<' || c=='>');c=nc());  for (;c=='<' || c=='>';s[++len]=c,c=nc());}const int N=105;const int P=1e9+7;int n;char s[N];struct edge{  int u,v,next;}G[N];int head[N],inum;inline void add(int u,int v,int p){  G[p].u=u; G[p].v=v; G[p].next=head[u]; head[u]=p;}#define V G[p].vll C[N][N],f[N][N];ll g[N];int size[N];inline void dfs(int u){  size[u]=1;  for (int p=head[u];p;p=G[p].next)    dfs(V);  f[u][1]=1;  for (int p=head[u];p;p=G[p].next){    int n=size[u],m=size[V];    for (int i=1;i<=n+m;i++) g[i]=0;    for (int i=1;i<=n;i++)      for (int j=1;j<=m;j++)    if (s[V]=='>')      for (int x=j;x<=m;x++)        g[x+i]+=f[u][i]*f[V][j]%P*C[i+x-1][x]%P*C[n+m-i-x][m-x]%P;    else      for (int x=0;x<j;x++)        g[x+i]+=f[u][i]*f[V][j]%P*C[i+x-1][x]%P*C[n+m-i-x][m-x]%P;    for (int i=1;i<=n+m;i++) f[u][i]=g[i]%P;    size[u]+=size[V];  }}int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(n); read(s+1);  for (int i=2;i<=n;i++) add(i/2,i,++inum);  C[0][0]=1; for (int i=1;i<=n;i++) { C[i][0]=1; for (int j=1;j<=i;j++) C[i][j]=(C[i-1][j]+C[i-1][j-1])%P; }  dfs(1);  ll ans=0;  for (int i=1;i<=n;i++)    ans+=f[1][i];  printf("%d\n",ans%P);  return 0;}

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