题意

sol

若某一方胜利，则设该方战胜的区间为 $[l_i,r_i]$，那么过程可描述为 $1$ 打败 $[l_1,r_1]$，$2$ 打败 $l_2,r_2$…………$k$ 打败 $[l_k,r_k]$。显然，$k$ 打败 $[l_k,r_k]$…………$2$ 打败 $l_2,r_2$，$1$ 打败 $[l_1,r_1]$ 与之等价，即顺序的正反对结果没有影响。
因此可以将序列翻转进行计算。设 $f_{u,j}$ 表示从第 $u$ 个点开始，会被上方的哪一个点击败，特别地，若全部战胜则 $f_{u,j}=0$。那么每次操作可以将序列倒序并进行处理，若最终结果为 $0$ 则胜利，否则失败。

代码

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

const int N = 200005, M = N * 2, K = 25;

int h[N], e[M], ne[M], idx;
int vs[K][K];
int n, m, q;
int c[N], f[N][K];
int w[N];

void add(int a, int b){
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}

void dfs_init(int u, int father){
    for (int i = 1; i <= m; i ++ ) 
        if (vs[i][c[u]]) 
            f[u][i] = f[father][i];
        else 
            f[u][i] = u;

    for (int i = h[u]; ~i; i = ne[i]){
        int j = e[i];
        if (j == father) continue;
        dfs_init(j, u);
    }
}

int main(){
    // freopen("ex_tree.in", "r", stdin);
    memset(h, -1, sizeof h);
    scanf("%d%d%d", &n, &m, &q);
    for (int i = 1; i <= m; i ++ )
        for (int j = 1; j <= m; j ++ )
            scanf("%d", &vs[i][j]);
    for (int i = 1; i <= n; i ++ ) scanf("%d", &c[i]);
    for (int i = 1; i < n; i ++ ){
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y), add(y, x);
    }

    dfs_init(1, 0);

    while (q -- ){
        int k, p;
        scanf("%d%d", &k, &p);
        for (int i = 1; i <= k; i ++ ) scanf("%d", &w[i]);
        for (int i = k; i; i -- ) p = f[p][w[i]];
        if (p) puts("0");
        else puts("1");
    }
}