Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P5546 [POI2000]公共串

P5546 [POI2000]公共串

对于第一个串建 sam ,记 $f _ i$ 表示节点 $i$ (显然是选择节点 $i$ 中最长的)和后面的串匹配的最大长度,对于每一个匹配,对当前点及其所有祖先都可以更新答案,那么最后统一一次dfs更新。最后所有串的答案是取最小值。

查看代码
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#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 2e5 + 10;
struct Node
{
int fa, len;
int nxt[26];
} tr[N];
char str[N];
int n, m, ans[N], last = 1, cnt = 1;
void insert(int c)
{
int p = last, np = ++cnt;
tr[np].len = tr[p].len + 1;
for (; p && !tr[p].nxt[c]; p = tr[p].fa)
tr[p].nxt[c] = np;
last = cnt;
if (!p)
{
tr[np].fa = 1;
return;
}
int q = tr[p].nxt[c];
if (tr[q].len == tr[p].len + 1)
{
tr[np].fa = q;
return;
}
int nq = ++cnt;
tr[nq] = tr[q], tr[nq].len = tr[p].len + 1;
tr[q].fa = tr[np].fa = nq;
for (; p && tr[p].nxt[c] == q; p = tr[p].fa)
tr[p].nxt[c] = nq;
}
int f[N];
int idx, hd[N], nxt[N], edg[N];
void add(int a, int b)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
}
void dfs(int x)
{
for (int i = hd[x]; ~i; i = nxt[i])
{
dfs(edg[i]);
f[x] = max(f[x], f[edg[i]]);
}
}
int main()
{
scanf("%d%s", &m, str + 1);
m--;
n = strlen(str + 1);
for (int i = 1; i <= n; i++)
insert(str[i] - 'a');
for (int i = 1; i <= cnt; i++)
ans[i] = tr[i].len;
for (int i = 1; i <= cnt; i++)
hd[i] = -1;
for (int i = 2; i <= cnt; i++)
add(tr[i].fa, i);
while (m--)
{
scanf("%s", str + 1);
n = strlen(str + 1);
for (int i = 1; i <= cnt; i++)
f[i] = 0;
int p = 1, t = 0;
for (int i = 1; i <= n; i++)
{
int c = str[i] - 'a';
while (p > 1 && !tr[p].nxt[c])
{
p = tr[p].fa;
t = tr[p].len;
}
if (tr[p].nxt[c])
{
p = tr[p].nxt[c];
t++;
}
f[p] = max(f[p], t);
}
dfs(1);
for (int i = 1; i <= cnt; i++)
ans[i] = min(ans[i], f[i]);
}
int res = 0;
for (int i = 1; i <= cnt; i++)
res = max(res, ans[i]);
printf("%d", res);
return 0;
}