学完了这阴间玩意肯定得知道这玩意怎么用吧
sam理论上可以
- 子串endpos大小
- 子串endpod位置
- 在后缀link上跑dp
- 在自动机上跑dp
- 统计有多少个子串经过一个点
- 二分
然后就是题目呗
P3975 [TJOI2015]弦论
给定字符串，求找到字典序第k大的子串
有两种询问：相同子串就算1个和每个子串都算一个的
首先把endpos大小跑出来
如果相同子串只算1个，那么子串到每个状态的贡献只有1，不然就是endpos大小。endpos大小可以把状态按len排序然后从大到小更新就行。
然后跑一下dfs，统计从点i开始的有多少个字符串
最后按字典序从自动机上跑就行
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| #include<iostream>
#include<map>
#include<stdio.h>
#include<cstring>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll MAXN =6e5 + 500;
string s;
ll k, t;
struct NODE {
int ch[26] = {};
ll fa = 0, len = 0;
}sam[MAXN << 1];
ll lst = 1, tot = 1;
ll f[MAXN << 1];
bool vis[MAXN << 1];
ll sz[MAXN << 1];
ll A[MAXN << 1];
void add(ll c) {
ll p = lst;
ll np = lst = ++tot;
f[np] = 1;
sam[np].len = sam[p].len + 1;
for (; p && !sam[p].ch[c]; p = sam[p].fa)sam[p].ch[c] = np;
if (!p)sam[np].fa = 1;
else {
ll q = sam[p].ch[c];
if (sam[q].len == sam[p].len + 1)sam[np].fa = q;
else {
ll nq = ++tot; sam[nq] = sam[q];
sam[nq].len = sam[p].len + 1;
sam[q].fa = sam[np].fa = nq;
for (; p && sam[p].ch[c] == q; p = sam[p].fa)sam[p].ch[c] = nq;
}
}
}
ll dfs(ll p) {
if (vis[p])return sz[p];
vis[p] = 1;
ll tmp = f[p];
for (int i = 0; i < 26; i++)
if (sam[p].ch[i])
tmp += dfs(sam[p].ch[i]);
return sz[p] = tmp;
}
void solve(ll p) {
if (k > sz[p]) {
printf("%d", -1);
return;
}
if (k <= f[p])return;
k -= f[p];
for (int i = 0; i < 26; i++) {
if (!sam[p].ch[i])continue;
if (sz[sam[p].ch[i]] < k)k -= sz[sam[p].ch[i]];
else {
printf("%c", i + 'a');
solve(sam[p].ch[i]);
return;
}
}
}
bool cmp(ll &a, ll &b) {
return sam[a].len < sam[b].len;
}
int main() {
cin >> s;
for (int i = 0; i < s.length(); i++) {
add(ll(s[i]) - 'a');
}
scanf("%lld %lld", &t, &k);
if (t == 0)
for (int i = 2; i <= tot; i++)f[i] = 1;
else {
for (int i = 0; i <= tot; i++)A[i] = i;
sort(A + 1, A + tot + 1, cmp);
for (int i = tot; i > 0; i--)f[sam[A[i]].fa] += f[A[i]];
}
f[1]=f[0] = 0;
dfs(1);
solve(1);
}
|
SPOJ1811
求两个字符串里最长公共子串长度
把s建成后缀自动机
然后把t的前缀挨个加进去，其实就是看t的前缀的后缀能在s里最长是多少
假设当前t加入的字符是c，当前状态是v，若v有c的出边，ans直接++
如果没有，就一直找v的后缀link到有c的出边为止，ans=len+1了
如果这玩意到了0状态，就把他设回源点状态，ans=0
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| #include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<string>
using namespace std;
typedef long long ll;
const ll MAXN = 5e5 + 5;
struct NODE{
int ch[26] = {};
int len, fa;
}sam[MAXN << 1];
ll lst = 1, tot = 1;
string s1, s2;
void add(ll c) {
ll p = lst; ll np = lst = ++tot;
sam[np].len = sam[p].len + 1;
for (; p && !sam[p].ch[c]; p = sam[p].fa)sam[p].ch[c] = np;
if (!p)sam[np].fa = 1;
else {
ll q = sam[p].ch[c];
if (sam[q].len == sam[p].len + 1)sam[np].fa = q;
else {
ll nq = ++tot; sam[nq] = sam[q];
sam[nq].len = sam[p].len + 1;
sam[q].fa = sam[np].fa = nq;
for (; p && sam[p].ch[c] == q; p = sam[p].fa)sam[p].ch[c] = nq;
}
}
}
int main() {
cin >> s1 >> s2;
ll ans = 0;
for (int i = 0; i < s1.length(); i++)add(s1[i] - 'a');
ll v = 1, l = 0;;
for (int i = 0; i < s2.length(); i++) {
ll tmp = s2[i] - 'a';
if (sam[v].ch[tmp]) {
v = sam[v].ch[tmp];
l++;
}
else {
for (; v && !sam[v].ch[tmp]; v = sam[v].fa);
if (v == 0) {
l = 0, v = 1;
}
else {
l = sam[v].len + 1;
v = sam[v].ch[tmp];
}
}
ans = max(ans, l);
}
cout << ans;
}
|
