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。

#include <bits/stdc++.h>

const int N = 110, MO = 1000000007;

struct Edge {
    int nex, v, len; /// 0  fa<son    1 fa>son
}edge[N << 1]; int tp;

int f[N][N], e[N], n, siz[N], temp[N], C[N][N];
char str[N];

inline void add(int x, int y, int z) {
    tp++;
    edge[tp].v = y;
    edge[tp].len = z;
    edge[tp].nex = e[x];
    e[x] = tp;
    return;
}

void DFS(int x) {
    siz[x] = 1;
    f[x][1] = 1;
    for(int i = e[x]; i; i = edge[i].nex) {
        int y = edge[i].v;
        //printf("%d -> %d 
", x, y);
        DFS(y);
        /// DP
        memcpy(temp, f[x], sizeof(f[x]));
        memset(f[x], 0, sizeof(f[x]));
        for(int j = 1; j <= siz[x]; j++) {
            /// temp[j]
            if(!temp[j]) continue;
            for(int k = 1; k <= siz[y]; k++) {
                //if(y == 5) printf("len = %d 
", edge[i].len);
                if(edge[i].len) { /// fa > son
                    for(int p = j + k; p <= j + siz[y]; p++) {
                        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;
                        //printf("f %d %d = %d 
", x, p, f[x][p]);
                    }
                }
                else { /// fa < son
                    for(int p = j; p <= j + k - 1; p++) {
                        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;
                        //printf("f %d %d = %d 
", x, p, f[x][p]);
                    }
                }
            }
        }
        siz[x] += siz[y];
    }
    return;
}

int main() {
    scanf("%d", &n);
    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]) % MO;
        }
    }
    scanf("%s", str);
    for(int i = 2; i <= n; i++) {
        if(str[i - 2] == '<') {
            add(i / 2, i, 0);
        }
        else {
            add(i / 2, i, 1);
        }
    }

    DFS(1);

    int ans = 0;
    for(int i = 1; i <= n; i++) {
        ans = (ans + f[1][i]) % MO;
    }
    printf("%d
", ans);
    return 0;
}