题目链接
题解
观察出题目中的关系实际上是完全二叉树的父子关系
我们设(f[i][j])为以(i)为根的节点在其子树中排名为(j)的方案数
转移时，枚举左右子树分别有几个节点比(i)小，进行转移
乍一看是(O(n^3))的，但其复杂度分析和某一题很像
就是在根处枚举两个子树大小，实质上就等于枚举任意两点(lca)，是(O(n^2))的
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define ls (u << 1)
#define rs (u << 1 | 1)
using namespace std;
const int maxn = 205,maxm = 100005,INF = 1000000000,P = 1000000007;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
LL C[maxn][maxn],f[maxn][maxn],siz[maxn],n,typ[maxn];
LL suml[maxn][maxn],sumr[maxn][maxn];
char s[maxn];
void init(){
for (int i = 0; i <= 100; i++){
C[i][0] = C[i][i] = 1;
for (int j = 1; j <= (i >> 1); j++)
C[i][j] = C[i][i - j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
}
}
void dfs(int u){
if (u > n) return;
dfs(ls); dfs(rs);
siz[u] = siz[ls] + 1 + siz[rs];
if (ls > n) f[u][1] = 1;
else if (rs > n){
for (int i = 0; i <= siz[ls]; i++)
if (!typ[ls]) f[u][i + 1] = sumr[ls][i + 1];
else f[u][i + 1] = suml[ls][i];
}
else {
LL t1,t2;
for (int i = 0; i <= siz[ls]; i++){
if (!typ[ls]) t1 = sumr[ls][i + 1];
else t1 = suml[ls][i];
for (int j = 0; j <= siz[rs]; j++){
if (!typ[rs]) t2 = sumr[rs][j + 1];
else t2 = suml[rs][j];
f[u][i + j + 1] = (f[u][i + j + 1] + C[i + j][i] * C[siz[u] - (i + j + 1)][siz[ls] - i] % P * t1 % P * t2 % P) % P;
}
}
}
for (int i = 1; i <= n ; i++) suml[u][i] = (suml[u][i - 1] + f[u][i]) % P;
for (int i = n; i >= 0; i--) sumr[u][i] = (sumr[u][i + 1] + f[u][i]) % P;
}
int main(){
init();
n = read();
scanf("%s",s + 2);
for (int i = 2; i <= n; i++)
typ[i] = s[i] == '<' ? 0 : 1;
dfs(1);
printf("%lld
",suml[1][n]);
return 0;
}