[POJ 2774]Long Long Message
阿新 • • 發佈:2018-07-07
math subset printf += get() std http namespace des
Description
題庫鏈接
給定兩個字符串 \(A\) 和 \(B\) ,求最長公共子串。
\(1\leq |A|,|B|\leq 100000\)
Solution
把串並起來求 \(height_i\) 的最大值,其中 \([suff(sa_{i-1})\subseteq B]\oplus [suff(sa_i)\subseteq B]=1\) 。
Code
#include <bits/stdc++.h> using namespace std; const int N = (100000+5)<<1; char ch[N]; int s1, n, m, x[N<<1], y[N<<1], c[N], sa[N], rk[N], height[N]; void get() { for (int i = 1; i <= n; i++) c[x[i] = ch[i]]++; for (int i = 2; i <= m; i++) c[i] += c[i-1]; for (int i = n; i >= 1; i--) sa[c[x[i]]--] = i; for (int k = 1; k <= n; k <<= 1) { int num = 0; for (int i = n-k+1; i <= n; i++) y[++num] = i; for (int i = 1; i <= n; i++) if (sa[i] > k) y[++num] = sa[i]-k; for (int i = 1; i <= m; i++) c[i] = 0; for (int i = 1; i <= n; i++) c[x[i]]++; for (int i = 2; i <= m; i++) c[i] += c[i-1]; for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i]; swap(x, y); x[sa[1]] = num = 1; for (int i = 2; i <= n; i++) x[sa[i]] = (y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k]) ? num : ++num; if ((m = num) == n) break; } for (int i = 1; i <= n; i++) rk[sa[i]] = i; for (int i = 1, k = 0; i <= n; i++) { if (rk[i] == 1) continue; if (k) --k; int j = sa[rk[i]-1]; while (i+k <= n && j+k <= n && ch[i+k] == ch[j+k]) ++k; height[rk[i]] = k; } } void work() { scanf("%s", ch+1); s1 = strlen(ch+1); ch[s1+1] = '$'; scanf("%s", ch+s1+2); n = strlen(ch+1); m = 255; get(); int ans = 0; for (int i = 2; i <= n; i++) if (((sa[i-1] <= s1)^(sa[i] <= s1)) && !(sa[i-1] == s1+1) && !(sa[i] == s1+1)) ans = max(ans, height[i]); printf("%d\n", ans); } int main() {work(); return 0; }
[POJ 2774]Long Long Message