洛谷-P3809-后缀排序(后缀数组)

看了求后缀数组的倍增法之后很快就理解了，但是自己写的倍增法用map排序还是超时了。然后看了两天别人写的模板，题目是通过了，但感觉代码还是半懂半背的。以后多熟悉熟悉吧；

• 后缀数组

  #include "bits/stdc++.h"
  using namespace std;
  const int MAXN = 1e6 + 5;
  char s[MAXN];
  int sa[MAXN], rk[MAXN], height[MAXN];
  void getSa(char* s, int* sa, int n, int m) {
      int* x = (int*)calloc(MAXN, sizeof(int));
      int* y = (int*)calloc(MAXN, sizeof(int));
      int* cnt = (int*)calloc(MAXN, sizeof(int));
      for (int i = 1; i <= n; i++) {
          cnt[x[i] = s[i]]++;
      }
      for (int i = 2; i <= m; i++) {
          cnt[i] += cnt[i - 1];
      }
      for (int i = n; i; i--) {
          sa[cnt[x[i]]--] = i;
      }
      for (int j = 1; j < n; j <<= 1) {
          int num = 0;
          for (int i = n - j + 1; i <= n; i++) {
              y[++num] = i;
          }
          for (int i = 1; i <= n; i++) {
              if (sa[i] > j) {
                  y[++num] = sa[i] - j;
              }
          }
          for (int i = 1; i <= m; i++) {
              cnt[i] = 0;
          }
          for (int i = 1; i <= n; i++) {
              cnt[x[i]]++;
          }
          for (int i = 2; i <= m; i++) {
              cnt[i] += cnt[i - 1];
          }
          for (int i = n; i; i--) {
              sa[cnt[x[y[i]]]--] = y[i];
          }
          swap(x, y);
          x[sa[1]] = num = 1;
          for (int i = 2; i <= n; i++) {
              if (y[sa[i]] != y[sa[i - 1]] || y[sa[i] + j] != y[sa[i - 1] + j]) {
                  x[sa[i]] = ++num;
              } else {
                  x[sa[i]] = num;
              }
          }
          if (num >= n) {
              break;
          }
          m = num;
      }
      free(x);
      free(y);
      free(cnt);
  }
  void getHeight(char* s, int* sa, int* rk, int* height, int n) {
      for (int i = 1; i <= n; i++) {
          rk[sa[i]] = i;
      }
      int k = 0;
      for (int i = 1; i <= n; i++) {
          k = k != 0 ? k - 1 : k;
          int j = sa[rk[i] - 1];
          while (s[i + k] == s[j + k]) {
              k++;
          }
          height[rk[i]] = k;
      }
  }
  int main() {
      gets(s + 1);
      int len = strlen(s + 1);
      getSa(s, sa, len, 130);
      // getHeight(s, sa, rk, height, len);
      for (int i = 1; i < len; i++) {
          printf("%d ", sa[i]);
      }
      printf("%d\n", sa[len]);
      /*
      for (int i = 1; i < len; i++) {
          printf("%d ", rk[i]);
      }
      printf("%d\n", rk[len]);
      for (int i = 1; i < len; i++) {
          printf("%d ", height[i]);
      }
      printf("%d\n", height[len]);
      */
      return 0;
  }

  附带rank数组和height数组，本题不需要。