题意：
给定n deep
1、构造一个n个节点的带权树，且最大深度为deep，每一个节点最多仅仅能有2个儿子
2、每一个节点的值为2^0, 2^1 ··· 2^(n-1) 随意两个节点值不能同样
3、对于一个节点，若他有左右儿子，则左子树的和 < 右子树的和
问：
有多少种构造方法。
思路：
dp
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
ret*=sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x <0) {
putchar('-');
x = -x;
}
if(x>9) pt(x/10);
putchar(x%10+'0');
}
///////////////////////////////////
typedef long long ll;
const int N = 362;
const ll mod = 1000000000 + 7;
int d[N][N], p[N][N], c[N][N];
int main() {
memset(d, 0, sizeof d);
memset(p, 0, sizeof p);
memset(c, 0, sizeof c);
for (int i = 0; i < N; ++i) {
c[i][0] = c[i][i] = 1;
for (int j = 1; j < i; ++j)
c[i][j] = (c[i - 1][j] + c[i-1][j-1]) % mod;
}
for (int i = 0; i < N; ++i)
p[0][i] = 1;
for (int i = 1; i < N; ++i)
p[1][i] = 1;
for (int i = 2; i < N; ++i)
for (int j = 1; j < N; ++j) {
p[i][j] += (ll)p[i-1][j-1] * 2 % mod;
p[i][j] %= mod;
if (i - 1 >= 2) {
for (int k = 1; k <= i - 2; ++k) {
p[i][j] += ((ll)c[i-2][k] * p[k][j-1] % mod) * p[i-1-k][j-1] % mod;
p[i][j] %= mod;
}
}
p[i][j] = (ll)p[i][j] * i % mod;
}
//int x, y; while (~scanf("%d%d", &x, &y)) printf("%d
", p[x][y]);
d[1][1] = 1; d[0][0] = 1;
for (int i = 2; i < N; ++i)
for (int j = i; j < N; ++j) {
d[i][j] += (ll)2 * d[i-1][j-1] % mod;
d[i][j] %= mod;
if (j - 1 >= 2) {
for (int k = 1; k <= j - 2; ++k) {
d[i][j] += ((ll)c[j-2][k] * p[k][i-2] % mod) * d[i-1][j-1-k] % mod;
d[i][j] %= mod;
d[i][j] += ((ll)c[j-2][k] * d[i-1][k] % mod) * p[j-1-k][i-2] % mod;
d[i][j] %= mod;
d[i][j] += ((ll)c[j-2][k] * d[i-1][k] % mod) * d[i-1][j-1-k] % mod;
d[i][j] %= mod;
}
}
d[i][j] = (ll)d[i][j] * j % mod;
}
int T = 0, cas, u, v;
scanf("%d", &cas);
while (cas -- > 0) {
rd(u); rd(v);
printf("Case #");
pt(++T);
putchar(':');
putchar(' ');
pt(d[v][u]);
putchar('
');
}
return 0;
}