PAT (Top Level) Practice1005 Programming Pattern (35 分)
阿新 • • 發佈:2018-12-23
題目連結:https://pintia.cn/problem-sets/994805148990160896/problems/994805154748940288
用字尾陣列求出長度為n所有子字串的大小關係,然後得出重複最多的。
演算法:字尾陣列
#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<set> #define maxn 1100000 using namespace std; int n; int sa[maxn], ranks[maxn]; int height[maxn]; int t1[maxn], t2[maxn]; int c[maxn]; string s; int m; bool cmp(int *r, int a, int b, int l) { return r[a] == r[b] && r[a + l] == r[b + l]; } void da(int t) { int p; m = s.length(); int *x = t1, *y = t2; for (int i = 0; i < t; i++) { c[i] = 0; } for (int i = 0; i < m; i++) { c[x[i] = s[i]]++; } for (int i = 1; i < t; i++) { c[i] += c[i - 1]; } for (int i = m - 1; i >= 0; i--) { sa[--c[x[i]]] = i; } for (int k = 1; k <= m;) { p = 0; for (int i = m - k; i < m; i++) { y[p++] = i; } for (int i = 0; i < m; i++) { if (sa[i] >= k) { y[p++] = sa[i] - k; } } for (int i = 0; i < t; i++) { c[i] = 0; } for (int i = 0; i < m; i++) { c[x[y[i]]]++; } for (int i = 1; i < t; i++) { c[i] += c[i - 1]; } for (int i = m - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i]; swap(x, y); p = 1; x[sa[0]] = 0; for (int i = 1; i < m; i++) { x[sa[i]] = cmp(y, sa[i - 1], sa[i], k) ? p - 1 : p++; } if (p >= m) break; t = p; k = k << 1; if (k >= n) { for (int i = 0; i < m; i++) { ranks[sa[i]] = i; } break; } } } void getheight() { int k = 0; for (int i = 0; i < m; height[ranks[i++]] = k) { if (k)--k; if (ranks[i] == 0) { height[ranks[i]] = 0; continue; } for (int j = sa[ranks[i] - 1]; s[i + k] == s[j + k];) { k++; if (k >= n) { k = n; break; } } } } int main() { cin >> n; cin.get(); getline(cin, s); da(128); getheight(); int h = sa[0], _h = -1; int ans = 1, ans1 = 1; for (int i = 1; i < m; i++) { if (height[i] >= n) { ans++; } else { if (ans > ans1||(_h==-1)||(ans==ans1&&s[h]<s[_h])) { ans1 = ans; _h = h; } ans = 1, h = sa[i]; } } if (ans > ans1 || (ans == ans1 && s[h] < s[_h])) { ans1 = ans; _h = h; } cout << s.substr(_h, n) << " " << ans1 << endl; return 0; }