[POJ 2774]Long Long Message

标签：problem   desc   const   tin   clu   oid   rip   .net   using   

Description

题库链接

给定两个字符串 \(A\) 和 \(B\) ，求最长公共子串。

\(1\leq |A|,|B|\leq 100000\)

Solution

把串并起来求 \(height_i\) 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

标签：problem   desc   const   tin   clu   oid   rip   .net   using   

原文地址：https://www.cnblogs.com/NaVi-Awson/p/9276300.html