[POI2000]パブリック文字列 - 接尾辞配列

説明

文字列の最長共通部分の長さの数を探しています。

解決

すべてのこれらの作品を一緒文字列の数を考慮し、文字セット番号の真ん中の記号が分離されていません。

二分法次いで答え\（K \）を連続期間場合、\（高さ\）以上である（K \）\、それが可能であり、一旦各ストリングは、少なくとも現れます。

コード

#include <bits/stdc++.h>
using namespace std;

const int _ = 1e5 + 10;
int N, n, s[_], belong[_];
int rnk[_], sa[_], height[_];
char str[_];

void SA() {
    static int t[_], a[_], buc[_], fir[_], sec[_], tmp[_];
    copy(s + 1, s + N + 1, t + 1);
    sort(t + 1, t + N + 1);
    int *end = unique(t + 1, t + N + 1);
    for (int i = 1; i <= N; ++i) a[i] = lower_bound(t + 1, end, s[i]) - t;
    fill(buc + 1, buc + N + 1, 0);
    for (int i = 1; i <= N; ++i) ++buc[a[i]];
    for (int i = 1; i <= N; ++i) buc[i] += buc[i - 1];
    for (int i = 1; i <= N; ++i) rnk[i] = buc[a[i] - 1] + 1;
    for (int len = 1; len <= N; len <<= 1) {
        for (int i = 1; i <= N; ++i) {
            fir[i] = rnk[i];
            sec[i] = i + len > N ? 0 : rnk[i + len];
        }
        fill(buc + 1, buc + N + 1, 0);
        for (int i = 1; i <= N; ++i) ++buc[sec[i]];
        for (int i = 1; i <= N; ++i) buc[i] += buc[i - 1];
        for (int i = 1; i <= N; ++i) tmp[N - --buc[sec[i]]] = i;
        fill(buc + 1, buc + N + 1, 0);
        for (int i = 1; i <= N; ++i) ++buc[fir[i]];
        for (int i = 1; i <= N; ++i) buc[i] += buc[i - 1];
        for (int i, j = 1; j <= N; ++j) {
            i = tmp[j];
            sa[buc[fir[i]]--] = i;
        }
        bool same = false;
        for (int i, j = 1, last = 0; j <= N; ++j) {
            i = sa[j];
            if (!last) rnk[i] = 1;
            else if (fir[i] == fir[last] && sec[i] == sec[last])
                rnk[i] = rnk[last], same = true;
            else rnk[i] = rnk[last] + 1;
            last = i;
        }
        if (!same) break;
    }
    for (int i = 1, k = 0; i <= N; ++i) {
        if (rnk[i] == 1) k = 0;
        else {
            if (k > 0) --k;
            int j = sa[rnk[i] - 1];
            while (i + k <= N && j + k <= N && a[i + k] == a[j + k]) ++k;
        }
        height[rnk[i]] = k;
    }
}


bool check(int k) {
    static int vis[_], tot = 0;
    int cnt = 0;
    ++tot;
    for (int i = 1; i <= N; ++i) {
        if (height[i] < k) cnt = 0, ++tot;
        else {
            if (vis[belong[sa[i]]] != tot)
                vis[belong[sa[i]]] = tot, ++cnt;
            if (vis[belong[sa[i - 1]]] != tot)
                vis[belong[sa[i - 1]]] = tot, ++cnt;
            if (cnt == n) return true;
        }
    }
    return false;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("string.in", "r", stdin);
    freopen("string.out", "w", stdout);
#endif
    scanf("%d", &n);
    int now = 0;
    for (int i = 1; i <= n; ++i) {
        ++now;
        scanf("%s", str);
        int len = strlen(str);
        for (int j = now; j <= now + len - 1; ++j)
            s[j] = str[j - now] - 'a' + 1, belong[j] = i;
        now += len - 1;
        s[++now] = i + 26;
    }
    N = now;
    SA();
    int l = 0, r = N;
    while (l < r) {
        int mid = (l + r + 1) >> 1;
        if (check(mid)) l = mid;
        else r = mid - 1;
    }
    printf("%d\n", l);
    return 0;
}