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  • 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
    ", 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;
    }
    
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  • 原文地址:https://www.cnblogs.com/dream-maker-yk/p/10071830.html
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