[HEOI2013] SAO

[HEOI2013] SAO

　　This question is a good question counts investigated swap the order of summation and then prefixes and optimization , design difficulties in the state, more study thinking.
　　A topic intended: to give you a number, there is a tree edge to edge, find the topological order.
　　The number of DAG topology is seeking a NP problem, but here is guaranteed to be a tree, so we can use a tree DP to solve.
　　The design condition, the light set node numbers \ (U \) is not enough, but also need to design a one-dimensional \ (I \) represents the junction \ (U \) in order to \ (U \) topological subtree rooted in sequences in the \ (i \) position, so we can write the transfer equation.
　　For \ (U \ rightarrow V \)
\ [F. '[U] [K] = \ Sigma_ {V \ in Son} F. [U] [I] \ Times F. [V] [J] \ Times \ tbinom {K -1} {i-1} \
times \ tbinom {size_u + size_v-k} {size_u-i}, i \ leq k \ leq i + j-1 \] 　　to \ (U \ LeftArrow V \)
\ [F. '[u] [k] = \ Sigma_ {v \ in son} F [u] [i] \ times F [v] [j] \ times \ tbinom {k-1} {i-1} \ times \ tbinom {size_u + size_v-k} {
size_u-i}, i + j \ leq k \ leq size_v + i \] 　　objectives:
\ [\ Sigma_. 1 = {I} ^ {N} F. [. 1] [I] \]
　　(all nodes of the standard automatic + 1)
　　found definitely not triple loop, the exchange \ (J \) and \ (K \) sequence found \ (F [v] [j ] \) can be processed and a prefix, a one-dimensional cut off.
　　In fact do so by the overall complexity of the \ (\ text {O} ( N ^ 3) \) reduced \ (\ text {O} (N ^ 2) \) . (Careful study of the following code is found in practice that the processing times of the number of points)
　　look at the details code.

#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define INF (1 << 30)
#define chkmax(a, b) a = max(a, b)
#define chkmin(a, b) a = min(a, b);

inline int read() {
    int w = 0, f = 1; char c;
    while (!isdigit(c = getchar())) f = c == '-' ? -1 : f;
    while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ 48), c = getchar();
    return w * f;
}

inline int read_ch() {
    char c;
    while (c = getchar(), c != '>' && c != '<');
    return c == '<';
}

const int maxn = 1000 + 5;
const int MOD = 1e9 + 7;

struct Edge {
    int v, w, pre;
} e[maxn << 1];
int m, G[maxn];
void clear() {
    m = 0;
    memset(G, -1, sizeof(G));
}
void add(int u, int v, int w) {
    e[m++] = (Edge){v, w, G[u]};
    G[u] = m-1;
}

int T, N;
int f[maxn][maxn], g[maxn], C[maxn][maxn];

void inc(int &a, int b) {
    a += b;
    if (a >= MOD) a -= MOD;
}

int dec(int a) {
    if (a < 0) a += MOD;
    return a;
}

void init() {
    C[0][0] = 1;
    for (register int i = 1; i <= 1000; i++) {
        C[i][0] = 1;
        for (register int j = 1; j <= i; j++)
            C[i][j] = (C[i-1][j] + C[i-1][j-1]) % MOD;
    }
}

int size[maxn];

void dfs(int u, int fa) {
    size[u] = 1;
    f[u][1] = 1;
    for (register int i = G[u]; ~i; i = e[i].pre) { \\ 这里实际上相当于u<->v之间的拓扑序合并起来
        int v = e[i].v;
        if (v == fa) continue;
        dfs(v, u);
        memcpy(g, f[u], sizeof(g));
        memset(f[u], 0, sizeof(f[u]));
        if (e[i].w) {
            for (register int i = 1; i <= size[u]; i++)
                for (register int k = i; k <= i+size[v]-1; k++)
                    inc(f[u][k], (ll)g[i] * dec(f[v][size[v]]-f[v][k-i]) % MOD * C[k-1][i-1] % MOD * C[size[u]+size[v]-k][size[u]-i] % MOD);
        } else {
            for (register int i = 1; i <= size[u]; i++)
                for (register int k = i+1; k <= size[v] + i; k++)
                    inc(f[u][k], (ll)g[i] * dec(f[v][k-i]) % MOD * C[k-1][i-1] % MOD * C[size[u]+size[v]-k][size[u]-i] % MOD);
        }
        size[u] += size[v];
    }
    for (register int i = 1; i <= size[u]; i++) inc(f[u][i], f[u][i-1]);
}

int main() {
    init();
    T = read();
    while (T--) {
        N = read();
        clear();
        for (register int i = 1; i < N; i++) {
            int u = read()+1, opt = read_ch(), v = read()+1;
            add(u, v, opt); add(v, u, !opt);
        }

        dfs(1, 1);

        printf("%d\n", f[1][N]);
    }

    return 0;
}