POJ 2217 Secretary [后缀数组]

最新推荐文章于 2020-05-26 15:06:03 发布

原创最新推荐文章于 2020-05-26 15:06:03 发布 · 220 阅读

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该博客聚焦于模板题，旨在求解两个字符串的最长公共子串，特别强调是子串而非子序列，采用后缀数组来实现，并推荐参考《挑战程序设计竞赛》P378 - 384的内容。

模板题
求两个字符串的最长公共子串
注意是子串，不是子序列
使用后缀数组来实现
看《挑战程序设计竞赛》P378~384

#define LOG(x) cout << #x << " = " << (x) << endl
#define PRINTLN(x) cout << (x) << endl
#define MEM(x) memset((x), 0, sizeof((x)))
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
int nextInt()
{
  int x;
  scanf("%d\n", &x);
  return x;
}
ll nextLL()
{
  ll x;
  scanf("%lld", &x);
  return x;
}
//TEMPLATE
int n, k;
const int MAXN = 10000 + 9;
int ranks[MAXN];
int tmp[MAXN];
int sa[MAXN];
int lcp[MAXN];
bool compare_sa(int i, int j)
{
  if (ranks[i] != ranks[j]) return ranks[i] < ranks[j];
  int ri = i + k <= n ? ranks[i + k] : -1;
  int rj = j + k <= n ? ranks[j + k] : -1;
  return ri < rj;
}
void construct_sa(string s, int *sa)
{
  n = s.size();
  for (int i = 0; i <= n; i++) {
    sa[i] = i;
    ranks[i] = i < n ? s[i] : -1;
  }
  for (k = 1; k <= n; k *= 2) {
    sort(sa, sa + n + 1, compare_sa);
    tmp[sa[0]] = 0;
    for (int i = 1; i <= n; i++) {
      tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
    }
    for (int i = 0; i <= n; i++) {
      ranks[i] = tmp[i];
    }
  }
}
void construct_lcp(string s, int *sa, int *lcp)
{
  int n = s.size();
  for (int i = 0; i <= n; i++) ranks[sa[i]] = i;
  int h = 0;
  lcp[0] = 0;
  for (int i = 0; i < n; i++) {
    int j = sa[ranks[i] - 1];
    if (h > 0) h--;
    for (; j + h < n && i + h < n; h++) {
      if (s[j + h] != s[i + h]) break;
    }
    lcp[ranks[i] - 1] = h;
  }
}
//MAIN
string nextLine()
{
  string s;
  getline(cin, s);
  return s;
}
void solve()
{
  string s = nextLine(), t = nextLine();
  int sl = s.size();
  s += '$' + t;
  construct_sa(s, sa);
  construct_lcp(s, sa, lcp);
  int ans = 0;
  for (int i = 0; i < s.size(); i++) {
    if ((sa[i] < sl) != (sa[i + 1] < sl)) {
      ans = max(ans, lcp[i]);
    }
  }
  printf("Nejdelsi spolecny retezec ma delku %d.\n", ans);
}
int main()
{
  freopen("in.txt", "r", stdin);
  int t = nextInt();
  while (t--) {
    solve();
  }
}