重复出现超过m次的最长的子串的最大下标 后缀数组或Hash+LCP UVA 12206 - Stammering Aliens

题目链接

题意：给定一个序列，求出出现次数大于m，长度最长的子串的最大下标

思路：hash大法或后缀数组，然后二分答案，每次利用hash值去找出最大下标即可。

代码：

#include <cstdio>  #include <cstring>  #include <algorithm>  using namespace std;    const int MAXLEN = 40005;    struct Suffix {        int s[MAXLEN];      int sa[MAXLEN], t[MAXLEN], t2[MAXLEN], c[MAXLEN], n;      int rank[MAXLEN], height[MAXLEN];        void build_sa(int m) {      n++;      int i, *x = t, *y = t2;      for (i = 0; i < m; i++) c[i] = 0;      for (i = 0; i < n; i++) c[x[i] = s[i]]++;      for (i = 1; i < m; i++) c[i] += c[i - 1];      for (i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;      for (int k = 1; k <= n; k <<= 1) {          int p = 0;          for (i = n - k; i < n; i++) y[p++] = i;          for (i = 0; i < n; i++) if (sa[i] >= k) y[p++] = sa[i] - k;          for (i = 0; i < m; i++) c[i] = 0;          for (i = 0; i < n; i++) c[x[y[i]]]++;          for (i = 0; i < m; i++) c[i] += c[i - 1];          for (i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];          swap(x, y);          p = 1; x[sa[0]] = 0;          for (i = 1; i < n; i++)          x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]) ? p - 1 : p++;          if (p >= n) break;          m = p;      }      n--;      }        void getHeight() {      int i, j, k = 0;      for (i = 1; i <= n; i++) rank[sa[i]] = i;      for (i = 0; i < n; i++) {          if (k) k--;          int j = sa[rank[i] - 1];          while (s[i + k] == s[j + k]) k++;          height[rank[i]] = k;      }      }  } gao;    const int N = 40005;    int m;  char str[N];    int judge(int x) {      int ans = -1;      for (int i = 1; i <= gao.n; i++) {      if (gao.n - gao.sa[i] < x) continue;      int Max = gao.sa[i], cnt = 1;      while (gao.height[i + 1] >= x && i < gao.n) {          Max = max(Max, gao.sa[i + 1]);          cnt++;          i++;      }      if (cnt >= m)          ans = max(ans, Max);      }      return ans;  }    void solve() {      if (judge(1) == -1) {      printf("none\n");      return;      }      int l = 1, r = gao.n - m + 2;      while (l < r) {      int mid = (l + r) / 2;      if (judge(mid) != -1) l = mid + 1;      else r = mid;      }      l--;      printf("%d %d\n", l, judge(l));  }    int main() {      while (~scanf("%d", &m) && m) {      scanf("%s", str);      int len = strlen(str);      for (int i = 0; i < len; i++)          gao.s[i] = str[i] - 'a' + 1;      gao.s[len] = 0;      gao.n = len;      gao.build_sa(27);      gao.getHeight();      solve();      }      return 0;  }  

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 40000 + 10;const int x = 123;int n, m, pos;unsigned long long H[maxn], xp[maxn];unsigned long long hash[maxn];int rank[maxn];int cmp(const int& a, const int& b) {  return hash[a] < hash[b] || (hash[a] == hash[b] && a < b);}int possible(int L) {  int c = 0;  pos = -1;  for(int i = 0; i < n-L+1; i++) {    rank[i] = i;    hash[i] = H[i] - H[i+L]*xp[L];  }  sort(rank, rank+n-L+1, cmp);  for(int i = 0; i < n-L+1; i++) {    if(i == 0 || hash[rank[i]] != hash[rank[i-1]]) c = 0;    if(++c >= m) pos = max(pos, rank[i]);  }  return pos >= 0;}int main() {  char s[maxn];  while(scanf("%d", &m) == 1 && m) {    scanf("%s", s);    n = strlen(s);    H[n] = 0;    for(int i = n-1; i >= 0; i--) H[i] = H[i+1]*x + (s[i] - 'a');    xp[0] = 1;    for(int i = 1; i <= n; i++) xp[i] = xp[i-1]*x;    if(!possible(1)) printf("none\n");    else {      int L = 1, R = n+1;      while(R - L > 1) {        int M = L + (R-L)/2;        if(possible(M)) L = M; else R = M;      }      possible(L);      printf("%d %d\n", L, pos);    }  }  return 0;}