LOJ#3023 老C的键盘

给定树，每条边有个大于号或者小于号，表示两个节点编号的大小关系。问有多少种树满足条件。n <= 100

解：树形DP。

设f_ij表示以i为根的子树中i是第j小的。转移的时候要乘上两个组合数。

 1 #include <bits/stdc++.h>
 2 
 3 const int N = 110, MO = 1000000007;
 4 
 5 struct Edge {
 6     int nex, v, len; /// 0  fa<son    1 fa>son
 7 }edge[N << 1]; int tp;
 8 
 9 int f[N][N], e[N], n, siz[N], temp[N], C[N][N];
10 char str[N];
11 
12 inline void add(int x, int y, int z) {
13     tp++;
14     edge[tp].v = y;
15     edge[tp].len = z;
16     edge[tp].nex = e[x];
17     e[x] = tp;
18     return;
19 }
20 
21 void DFS(int x) {
22     siz[x] = 1;
23     f[x][1] = 1;
24     for(int i = e[x]; i; i = edge[i].nex) {
25         int y = edge[i].v;
26         //printf("%d -> %d \n", x, y);
27         DFS(y);
28         /// DP
29         memcpy(temp, f[x], sizeof(f[x]));
30         memset(f[x], 0, sizeof(f[x]));
31         for(int j = 1; j <= siz[x]; j++) {
32             /// temp[j]
33             if(!temp[j]) continue;
34             for(int k = 1; k <= siz[y]; k++) {
35                 //if(y == 5) printf("len = %d \n", edge[i].len);
36                 if(edge[i].len) { /// fa > son
37                     for(int p = j + k; p <= j + siz[y]; p++) {
38                         f[x][p] = (f[x][p] + 1ll * temp[j] * f[y][k] % MO * C[p - 1][j - 1] % MO * C[siz[x] + siz[y] - p][siz[x] - j] % MO) % MO;
39                         //printf("f %d %d = %d \n", x, p, f[x][p]);
40                     }
41                 }
42                 else { /// fa < son
43                     for(int p = j; p <= j + k - 1; p++) {
44                         f[x][p] = (f[x][p] + 1ll * temp[j] * f[y][k] % MO * C[p - 1][j - 1] % MO * C[siz[x] + siz[y] - p][siz[x] - j] % MO) % MO;
45                         //printf("f %d %d = %d \n", x, p, f[x][p]);
46                     }
47                 }
48             }
49         }
50         siz[x] += siz[y];
51     }
52     return;
53 }
54 
55 int main() {
56     scanf("%d", &n);
57     for(int i = 0; i <= n; i++) {
58         C[i][0] = C[i][i] = 1;
59         for(int j = 1; j < i; j++) {
60             C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MO;
61         }
62     }
63     scanf("%s", str);
64     for(int i = 2; i <= n; i++) {
65         if(str[i - 2] == '<') {
66             add(i / 2, i, 0);
67         }
68         else {
69             add(i / 2, i, 1);
70         }
71     }
72 
73     DFS(1);
74 
75     int ans = 0;
76     for(int i = 1; i <= n; i++) {
77         ans = (ans + f[1][i]) % MO;
78     }
79     printf("%d\n", ans);
80     return 0;
81 }