HDU 6208 The Dominator of Strings 读入挂＋kmp / AC自动机

题目链接

题意

给定 n 个串，问是否存在一个串包含其它所有串。

读入的问题

The total length of strings in each case has the limit of 100000.
The limit is 30MB for the input file.

考虑将所有的串读到一整个串里，记录每个串在其中的开始位置和长度

注意：这种情况下，如果每个串末尾有

'\0'

，则开的长度不是 100000 而是 200000

法一：kmp 546ms

思路

找到最长的串作为主串，其它 n−1 个串作为模式串，跑 n−1 遍 kmp

Code

#include <bits/stdc++.h>
#define maxm 200010
const int BUF_SIZE = (int)1e4+10;
using namespace std;
int f[maxm];
char s[maxm], ans[maxm];
struct node {
int beg, len;
node(int _a=0, int _b=0) : beg(_a), len(_b) {}
}nd[maxm];
struct fastIO {
char buf[BUF_SIZE];
int cur;
FILE *in, *out;
fastIO() {
cur = BUF_SIZE;
in = stdin;
out = stdout;
}
inline char nextChar() {
if(cur == BUF_SIZE) {
fread(buf, BUF_SIZE, 1, in);
cur = 0;
}
return buf[cur++];
}
inline int nextInt() {
int x = 0;
char c = nextChar();
while(!('0' <= c && c <= '9')) c = nextChar();
while('0' <= c && c <= '9') {
x = x * 10 + c - '0';
c = nextChar();
}
return x;
}
inline int nextString(char* s) {
char c = nextChar();
int len = 0;
while(!('a' <= c && c <= 'z')) c = nextChar();
while('a' <= c && c <= 'z') s[len++] = c, c = nextChar();
return len;
}
inline void printChar(char ch) {
buf[cur++] = ch;
if (cur == BUF_SIZE) {
fwrite(buf, BUF_SIZE, 1, out);
cur = 0;
}
}
inline void printInt(int x) {
if (x >= 10) printInt(x / 10);
printChar(x % 10 + '0');
}
inline void close() {
if (cur > 0) {
fwrite(buf, cur, 1, out);
}
cur = 0;
}
} IO;
void getfail(char* P, int m) {
f[0] = f[1] = 0;
for (int i = 1; i < m; ++i) {
int j = f[i];
while (j && P[j] != P[i]) j = f[j];
f[i+1] = P[j] == P[i] ? j+1 : 0;
}
}
bool kmp(char* T, int n, char* P, int m) {
getfail(P, m);
int j = 0;
for (int i = 0; i < n; ++i) {
while (j && T[i] != P[j]) j = f[j];
if (T[i] == P[j]) ++j;
if (j == m) return true;
}
return false;
}
int kas = 0;
void work() {
int n;
n = IO.nextInt();
memset(s, 0, sizeof(s));
int mx = -1, p = -1, cur = 0;
for (int i = 0; i < n; ++i) {
int temp = cur;
int len = IO.nextString(s+cur);
nd[i] = node(cur, len);
if (len > mx) mx = len, p = i;
cur += len+1;
}
for (int i = 0; i < n; ++i) {
if (p == i) continue;
if (!kmp(s+nd[p].beg, nd[p].len, s+nd[i].beg, nd[i].len)) { printf("No\n"); return; }
}
printf("%s\n", s+nd[p].beg);
}

int main() {
int T;
T = IO.nextInt();
while (T--) work();
return 0;
}

法二：AC自动机 2121ms

思路

对其他 n−1 个串建 AC自动机，拿最长的串在上面跑一遍。

Code

#include <bits/stdc++.h>
#define maxm 100010
const int BUF_SIZE = (int)1e4+10;
using namespace std;
int son[maxm][26], fail[maxm], cnt[maxm], tot;
char s[maxm * 2];
struct node {
int beg, len;
node(int _a=0, int _b=0) : beg(_a), len(_b) {}
}nd[maxm];
struct fastIO {
char buf[BUF_SIZE];
int cur;
FILE *in, *out;
fastIO() {
cur = BUF_SIZE;
in = stdin;
out = stdout;
}
inline char nextChar() {
if(cur == BUF_SIZE) {
fread(buf, BUF_SIZE, 1, in);
cur = 0;
}
return buf[cur++];
}
inline int nextInt() {
int x = 0;
char c = nextChar();
while(!('0' <= c && c <= '9')) c = nextChar();
while('0' <= c && c <= '9') {
x = x * 10 + c - '0';
c = nextChar();
}
return x;
}
inline int nextString(char* s) {
char c = nextChar();
int len = 0;
while(!('a' <= c && c <= 'z')) c = nextChar();
while('a' <= c && c <= 'z') s[len++] = c, c = nextChar();
return len;
}
inline void printChar(char ch) {
buf[cur++] = ch;
if (cur == BUF_SIZE) {
fwrite(buf, BUF_SIZE, 1, out);
cur = 0;
}
}
inline void printInt(int x) {
if (x >= 10) printInt(x / 10);
printChar(x % 10 + '0');
}
inline void close() {
if (cur > 0) {
fwrite(buf, cur, 1, out);
}
cur = 0;
}
} IO;
void add(char* s, int len) {
int p = 0;
for (int i = 0; i < len; ++i) {
if (son[p][s[i]-'a'] == -1) {
cnt[++tot] = 0;
for (int j = 0; j < 26; ++j) son[tot][j] = -1;
son[p][s[i]-'a'] = tot;
}
p = son[p][s[i]-'a'];
}
++cnt[p];
}
void build() {
queue<int> que;
fail[0] = 0;
for (int i = 0; i < 26; ++i) {
if (son[0][i] == -1) son[0][i] = 0;
else {
fail[son[0][i]] = 0;
que.push(son[0][i]);
}
}
while (!que.empty()) {
int x = que.front(); que.pop();
for (int i = 0; i < 26; ++i) {
if (son[x][i] == -1) son[x][i] = son[fail[x]][i];
else {
fail[son[x][i]] = son[fail[x]][i];
que.push(son[x][i]);
}
}
}
}
int query(char* s, int len) {
int p = 0, ret = 0;
for (int i = 0; i < len; ++i) {
p = son[p][s[i]-'a'];
int temp = p;
while (temp) {
if (cnt[temp] == -1) break;
ret += cnt[temp];
cnt[temp] = -1;
temp = fail[temp];
}
}
return ret;
}
int kas = 0;
void init() {
memset(s, 0, sizeof(s)); cnt[tot = 0] = 0;
for (int i = 0; i < 26; ++i) son[0][i] = -1;
}
void work() {
init();
int n;
n = IO.nextInt();
int mx = -1, p = -1, cur = 0;
for (int i = 0; i < n; ++i) {
int temp = cur;
int len = IO.nextString(s+cur);
nd[i] = node(cur, len);
if (len > mx) mx = len, p = i;
cur += len+1;
add(s+nd[i].beg, nd[i].len);
}
build();
if (query(s+nd[p].beg, nd[p].len) == n) printf("%s\n", s+nd[p].beg);
else printf("No\n");
}

int main() {
int T;
T = IO.nextInt();
while (T--) work();
return 0;
}