[POJ 1743]Musical Theme

Description

题库链接

给定一个长度为 \(n\) 的字符串，求最长重复子串，这两个子串不能重叠。（题目模型需转换）

\(1\leq n\leq 20000\)

Solution

先二分答案，把题目变成判定性问题：判断是否存在两个长度为 \(k\) 的子串是相同的，且不重叠。

解决这个问题的关键还是利用 \(height\) 数组。把排序后的后缀分成若干组，其中每组的后缀之间的 \(height\) 值都不小于 \(k\) 。

容易看出，有希望成为最长公共前缀不小于 \(k\) 的两个后缀一定在同一组。然后对于每组后缀，只须判断每个后缀的 \(sa\) 值的最大值和最小值之差是否不小于 \(k\) 。如果有一组满足，则说明存在，否则不存在。

因为后缀是有序的，相邻的后缀间的 \(LCP\) 必定的极大的；接下来就找到每个组里后缀 \(sa\) 值最大和最小的，如果差值不小于 \(k\) 就成立，因为这样小下标的后缀沿着 \(LCP\) 下去走 \(k\) 步才不会盖到大下标的后缀。

Code

#include <bits/stdc++.h>
using namespace std;
const int N = 20000+5;

int ch[N], n, m, x[N<<1], y[N<<1], c[N], sa[N], rk[N], height[N];

void get() {
    for (int i = 1; i <= m; i++) c[i] = 0;
    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;
    }
}
bool judge(int x) {
    for (int i = 2, maxn = sa[1], minn = sa[1]; i <= n; i++) {
        if (height[i] >= x) maxn = max(maxn, sa[i]), minn = min(minn, sa[i]);
        if (height[i] < x || i == n) {
            if (maxn-minn > x) return true;
            maxn = minn = sa[i];
        }
    }
    return false;
}
void work() {
    while (~scanf("%d", &n) && n) {
        m = 88*2;
        for (int i = 1; i <= n; i++) scanf("%d", &ch[i]); --n;
        for (int i = 1; i <= n; i++) ch[i] = ch[i+1]-ch[i]+88;
        get(); int L = 4, R = n, ans = -1;
        while (L <= R) {
            int mid = (L+R)>>1;
            if (judge(mid)) ans = mid, L = mid+1;
            else R = mid-1;
        }
        printf("%d\n", ans+1);
    }
}
int main() {work(); return 0; }