[POI2000]公共串 - 后缀数组
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Description
求若干个串的最长的公共子串的长度。
Solution
考虑将这若干个串全部拼起来,中间用一些不在字符集内的符号隔开。
然后二分答案 (K),如果连续的一段 (height) 都大于等于 (K),且每个串都出现了至少一次,则是可行的。
Code
#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
", l);
return 0;
}
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