BZOJ3473: 字符串【后缀数组+思维】

Description

给定n个字符串,询问每个字符串有多少子串(不包括空串)是所有n个字符串中至少k个字符串的子串?

Input

第一行两个整数n,k。

接下来n行每行一个字符串。

Output

一行n个整数,第i个整数表示第i个字符串的答案。

Sample Input

3 1
abc
a
ab

Sample Output

6 1 3

HINT

对于 100% 的数据,1<=n,k<=10^5,所有字符串总长不超过10^5,字符串只包含小写字母。


思路

首先发现这东西是真的不好做。。。那就找找性质

先把所有子串的贡献拆分成每个后缀的前缀的贡献

然后考虑怎么算每个后缀的贡献

又因为对于后缀i,假设前j个前缀有贡献,那么对于后缀i+1,它的前j-1个前缀一定是有贡献的

是不是就想到了height数组的处理,很容易证明这个的复杂度是线性的

那么怎么考虑可不可以扩展呢?

因为我们要算这个串的出现次数

并且知道连接所有串的分隔符一定是不会被匹配的

所以我们可以直接二分出包含这个串的排名区间

那么就可以预处理出排名第i的字符串至少要到第j个排名的字符串才能包含k个不同的字符串

这个东西可以双指针做


一定要把预处理的指针数组初值设成INF


 #include<bits/stdc++.h>
 
using namespace std;
 
typedef pair<int, int> pi;
typedef long long ll;
const int N = 2e5 + 10;
const int LOG = 20;
 
struct Suffix_Array {
  int s[N], n, m;
  int c[N], x[N], y[N];
  int height[N], sa[N], rank[N];
  int st[N][LOG], Log[N];
  ll sum[N]; 
   
  void init(int len, char *c) {
    n = len, m = 0;
    for (int i = 1; i <= len; i++) {
      s[i] = c[i];
      m = max(m, s[i]);
    }
  }
   
  void radix_sort() {
    for (int i = 1; i <= m; i++) c[i] = 0;
    for (int i = 1; i <= n; i++) c[x[y[i]]]++;
    for (int i = 1; i <= m; i++) c[i] += c[i - 1];
    for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];
  }
   
  void buildsa() {
    for (int i = 1; i <= n; i++) x[i] = s[i], y[i] = i;
    radix_sort();
    int now;
    for (int k = 1; k <= n; k <<= 1) {
      now = 0;
      for (int i = n - k + 1; i <= n; i++) y[++now] = i;
      for (int i = 1; i <= n; i++) if (sa[i] > k) y[++now] = sa[i] - k;
      radix_sort();
      y[sa[1]] = now = 1;
      for (int i = 2; i <= n; i++) y[sa[i]] = (x[sa[i]] == x[sa[i - 1]] && x[sa[i] + k] == x[sa[i - 1] + k]) ? now : ++now;
      swap(x, y);
      if (now == n) break;
      m = now;
    }
  }
   
  void buildrank() {
    for (int i = 1; i <= n; i++) rank[sa[i]] = i;
  }
   
  void buildsum() {
    for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + n - sa[i] + 1 - height[i];
  }
 
  void buildheight() {
    for (int i = 1; i <= n; i++) if (rank[i] != 1) {
      int k = max(height[rank[i - 1]] - 1, 0); 
      for (; s[i + k] == s[sa[rank[i] - 1] + k]; k++);
      height[rank[i]] = k;
    }
  }
   
  void buildst() {
    Log[1] = 0;
    for (int i = 2; i < N; i++) Log[i] = Log[i >> 1] + 1;
    for (int i = 1; i <= n; i++) st[i][0] = height[i];
    for (int j = 1; j < LOG; j++) {
      for (int i = 1; i + (1 << (j - 1)) <= n; i++) {
        st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
      }
    }
  }
   
  int queryst(int l, int r) {
    if (l == r) return n - sa[l] + 1;
    if (l > r) swap(l, r);
    ++l; //***
    int k = Log[r - l + 1];
    return min(st[l][k], st[r - (1 << k) + 1][k]);
  }
   
  int querylcp(int la, int lb) {
    return queryst(rank[la], rank[lb]);
  }
   
  int querylcp(int la, int ra, int lb, int rb) {
    return min(min(ra - la + 1, rb - lb + 1), queryst(rank[la], rank[lb]));
  }
   
  void build(int len, char *c) {
    init(len, c);
    buildsa();
    buildrank();
    buildheight();
    buildsum();
    buildst();
  } 
} Sa;
 
char s[N], c[N];
int len[N], bg[N], ed[N], tot = 0;
int n, k, rpos[N], num[N], bel[N];
 
bool check(int pos, int len) {
  int x = Sa.rank[pos], l, r;
  int l_line = x, r_line = x;
  l = 1, r = x - 1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (Sa.queryst(x, mid) > len) l_line = mid, r = mid - 1;
    else l = mid + 1;
  } 
  l = x + 1, r = tot;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (Sa.queryst(x, mid) > len) r_line = mid, l = mid + 1;
    else r = mid - 1;
  }
  return rpos[l_line] <= r_line;
}
 
int main() {
#ifdef dream_maker
  freopen("input.txt", "r", stdin);
#endif
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%s", c + 1);
    len[i] = strlen(c + 1);
    bg[i] = tot + 1;
    for (int j = 1; j <= len[i]; j++) {
      s[++tot] = c[j];
      bel[tot] = i;
    }
    ed[i] = tot;
    s[++tot] = '#';
  }
  Sa.build(tot, s);
  memset(rpos, 0x3f, sizeof(rpos)); //*****
  int l = 1, r = 0, cnt = 0;
  for (; r <= tot; r++) {
    if (!bel[Sa.sa[r]]) continue; 
    ++num[bel[Sa.sa[r]]];
    if (num[bel[Sa.sa[r]]] == 1) ++cnt;
    if (cnt >= k) {
      for (; l <= r; l++) {
        if (!bel[Sa.sa[l]]) continue;
        if (cnt >= k) rpos[l] = r;
        else break;
        if (num[bel[Sa.sa[l]]] == 1) --cnt;
        --num[bel[Sa.sa[l]]];
      }
    } 
  }
  for (int i = 1; i <= n; i++) {
    ll ans = 0; int cur = 0;
    for (int j = bg[i]; j <= ed[i]; j++) {
      cur = max(cur - 1, 0);
      while (j + cur <= ed[i] && check(j, cur)) ++cur;
      ans += cur;
    }
    printf("%lld ", ans);
  }
  return 0;
}
posted @ 2018-12-05 22:06 Dream_maker_yk 阅读(...) 评论(...) 编辑 收藏