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
