题意

求一个串中出现至少m次的子串的最大长度，对于最大长度，求出最大的左端点

题解

本来想练哈希的，没忍住就写了一个SAM

SAM拿来做就很裸了

只要检查每个节点的right集合大小是否不小于m，然后step[u]就表示u节点所代表字符串的最大长度

为了求出端点，我们还需要记录right集合的最大值

然后就水过啦

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<algorithm>

#define LL long long int

#define Redge(u) for (int k = h[u]; k; k = ed[k].nxt)

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define ULL unsigned long long int

#define cls(s) memset(s,0,sizeof(s))

using namespace std;

const int maxn = 80005,maxm = 100005,INF = 1000000000;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}

	return out * flag;

}

char s[maxn];

int m,n,b[maxn],a[maxn],ans,ansr;

int ch[maxn][26],pre[maxn],sz[maxn],l[maxn],r[maxn],last,cnt;

void init(){

	cls(ch); cls(pre); cls(sz); cls(l); cls(r); last = cnt = 1; ans = 0;

}

void ins(int x){

	int p = last,np = ++cnt; last = np;

	r[np] = l[np] = l[p] + 1; sz[np] = 1;

	while (p && !ch[p][x]) ch[p][x] = np,p = pre[p];

	if (!p) pre[np] = 1;

	else {

		int q = ch[p][x];

		if (l[q] == l[p] + 1) pre[np] = q;

		else {

			int nq = ++cnt;

			l[nq] = l[p] + 1;

			for (int i = 0; i < 26; i++) ch[nq][i] = ch[q][i];

			pre[nq] = pre[q]; pre[q] = pre[np] = nq;

			while (ch[p][x] == q) ch[p][x] = nq,p = pre[p];

		}

	}

}

void build(){

	for (int i = 1; i <= n; i++) ins(s[i] - 'a');

	for (int i = 0; i <= cnt; i++) b[i] = 0;

	for (int i = 1; i <= cnt; i++) b[l[i]]++;

	for (int i = 1; i <= cnt; i++) b[i] += b[i - 1];

	for (int i = 1; i <= cnt; i++) a[b[l[i]]--] = i;

}

void solve(){

	for (int i = cnt; i; i--){

		int u = a[i];

		sz[pre[u]] += sz[u];

		r[pre[u]] = max(r[pre[u]],r[u]);

		if (u != 1 && sz[u] >= m){

			if (l[u] > ans) ans = l[u],ansr = r[u] - l[u];

			else if (l[u] == ans && r[u] - l[u] > ansr) ansr = r[u] - l[u];

		}

	}

}

int main(){

	while (m = read()){

		init();

		scanf("%s",s + 1); n = strlen(s + 1);

		build();

		solve();

		if (ans) printf("%d %d\n",ans,ansr);

		else puts("none");

	}

	return 0;

}

二分 + hash的乱搞也写写