BZOJ1369/BZOJ2865 【后缀数组+线段树】

Description

XX在进行字符串研究的时候,遇到了一个十分棘手的问题。

在这个问题中,给定一个字符串S,与一个整数K,定义S的子串T=S(i, j)是关于第K位的识别子串,满足以下两个条件:

1、i≤K≤j。

2、子串T只在S中出现过一次。

例如,S="banana",K=5,则关于第K位的识别子串有"nana","anan","anana","nan","banan"和"banana"。

现在,给定S,XX希望知道对于S的每一位,最短的识别子串长度是多少,请你来帮助他。

Input

仅一行,输入长度为N的字符串S。

Output

输出N行,每行一个整数,第i行的整数表示对于第i位的最短识别子串长度。

Sample Input

agoodcookcooksgoodfood

Sample Output

1
2
3
3
2
2
3
3
2
2
3
3
2
1
2
3
3
2
1
2
3
4

HINT

N<=5*10^5


首先发现可以按照每个后缀统计贡献

然后直接把每个后缀和前后排名的串lcp的max:len求出来,这些值是不能更新的

然后对于$[i,i+len] \(用len+1更新,对于\)[i+len,n]$用一个等差数列更新min

等差数列维护直接标记永久化就可以了

然后注意特判是不是没有合法的解


#include<bits/stdc++.h>

using namespace std;

typedef pair<int, int> pi;
const int N = 5e5 + 10;
const int INF_of_int = 1e9;

int typ1[N << 2], typ2[N << 2];

#define LD (t << 1)
#define RD (t << 1 | 1)

void build(int t, int l, int r) {
  typ1[t] = INF_of_int;
  typ2[t] = -INF_of_int;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(LD, l, mid);
  build(RD, mid + 1, r);
}

void modify(int t, int l, int r, int ql, int qr, int typ, int val) {
  if (ql > qr) return;
  if (ql <= l && r <= qr) {
    if (typ == 1) {
      typ1[t] = min(typ1[t], val);
    } else {
      typ2[t] = max(typ2[t], val);
    }
    return;
  }
  int mid = (l + r) >> 1;
  if (qr <= mid) modify(LD, l, mid, ql, qr, typ, val);
  else if (ql > mid) modify(RD, mid + 1, r, ql, qr, typ, val);
  else {
    modify(LD, l, mid, ql, mid, typ, val);
    modify(RD, mid + 1, r, mid + 1, qr, typ, val);
  }
} 

void output(int t, int l, int r, int val, int pos) {
  if (typ1[t]) val = min(val, typ1[t]);
  if (typ2[t]) val = min(val, pos - typ2[t] + 1);
  if (l == r) {
    printf("%d\n", val);
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid) output(LD, l, mid, val, pos);
  else output(RD, mid + 1, r, val, pos); 
}

struct Suffix_Array {
  int s[N], n, m;
  int c[N], x[N], y[N];
  int sa[N], rank[N], height[N];
  
  void init(int len, char *c) {
    n = len, m = 0;
    for (int i = 1; i <= n; 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 buildheight() {
    for (int i = 1; i <= n; i++) {
      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 build(int len, char *c) {
    init(len, c);
    buildsa();
    buildrank();
    buildheight();
  } 
  
  void solve() {
    for (int i = 1; i <= n; i++) {
      int len = max(height[rank[i]], height[rank[i] + 1]);
      if (i + len > n) continue;
      modify(1, 1, n, i, i + len, 1, len + 1);
      modify(1, 1, n, i + len, n, 2, i);
    }
  }
} Sa;

int len;
char s[N];

int main() {
#ifdef dream_maker
  freopen("input.txt", "r", stdin);
#endif
  scanf("%s", s + 1);
  len = strlen(s + 1);
  Sa.build(len, s);
  build(1, 1, len);
  Sa.solve();
  for (int i = 1; i <= len; i++) output(1, 1, len, len, i);
  return 0;
}
posted @ 2018-12-05 16:46 Dream_maker_yk 阅读(...) 评论(...) 编辑 收藏