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#include#include #include #include #include #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