P2292，第二个点洛谷显示WA，本地显示RE

#include <cstdio>
#include <cstring>

using namespace std;

#define ll long long
#define R register
#define U unsigned
#define qwq printf("qwq\n")

const int N = 205, M = 2e6 + 5;

int n, m, tot = 1, len, ans;
int trie[N][30];
char s[M];
bool End[N];

inline int max (int a, int b) {return (a)>(b)?(a):(b);}

void insert (char a[]) {
	int len = strlen(a), p = 1;
	for (int i = 0; i < len; ++i) {
		//printf("%c\n", a[i]);
		int ch = a[i] - 'a';
		if (!trie[p][ch]) trie[p][ch] = ++tot;
		p = trie[p][ch];
	}
	//printf("\n");
	End[p] = 1;
}

void find (int pos, int p, int cnt) {
	printf("%d %d %d\n", pos, p, cnt);
	for (int i = pos; i < len; ++i) {
		if (pos == 88314 && p == 1 && cnt == 88314) qwq;
		if (pos == 88314 && p == 1 && cnt == 88314) printf("%d\n", trie[p][s[i] - 'a']);
		if (pos == 88314 && p == 1 && cnt == 88314) qwq;
		if (pos == 88314 && p == 1 && cnt == 88314) printf("%d\n", s[i] - 'a');
		p = trie[p][s[i] - 'a'];
		if (pos == 88314 && p == 1 && cnt == 88314) printf("%d\n", p);
		if (pos == 88314 && p == 1 && cnt == 88314) qwq;
		if (!p) return;
		if (pos == 88314 && p == 1 && cnt == 88314) qwq;
		++cnt;
		if (End[p]) {
			ans = max(ans, cnt);
			find(i + 1, 1, cnt);
			find(i + 1, p, cnt);
			return;
		}
	}
	ans = max(ans, cnt);
}

int main () {
	freopen("2.in", "r", stdin);
	//freopen("data.out", "w", stdout);
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) {
		char a[12];
		scanf("%s", a);
		insert(a);
	}
	for (int i = 1; i <= m; ++i) {
		scanf("%s", s);
		ans = 0;
		len = strlen(s);
		find(0, 1, 0);
		printf("%d\n", ans);
	}
	return 0;
}/*what is you your name
whatisyourname
使用单词what is your name时,可以理解上面句子
但是使用单词what is you name时无法理解
也就是说,如果trie当前节点有End标记且未结束时
要考虑回头或继续往下走*/

第二个测试点（提取码：1pwd）（第二个点太大了，剪贴板装不下）

运行结果：)

find函数里的前4个if都运行了，说明数组不可能越界，但是第5个if就没有运行了...求问哪儿出问题了qwq