hdu 5972 Regular Number 字符串Shift-And算法 + bitset

题目链接

题意

给定两个串(S,T)，找出(S)中所有与(T)匹配的子串。

这里，(T)的每位上可以有若干((leq 10))种选择，匹配的含义是：对于(S)的子串的每一位，(T)的相应位都有一种选择与之对应。

题解

shift-and算法详解 https://www.douban.com/note/321872072/

搜出来的题解全都是(shift-and)的...。

学习了一波...然而并不明白(kmp)为什么不可以...。

看到这道题直觉就是和hdu 4749一样美滋滋写了然后(T)了...。

Code

#include <bits/stdc++.h>
#define maxn 1010
#define maxm 5000010
using namespace std;
typedef long long LL;
char s[maxm];
bitset<maxn> B[11];
const int BUF_SIZE = (int)1e4+10;
namespace fastIO{
    #define BUF_SIZE 100000
    #define OUT_SIZE 1000000
    bool IOerror=0;
    inline char nc(){
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend){
            p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if (pend==p1){IOerror=1;return -1;}
        }return *p1++;
    }
    inline bool blank(char ch){return ch==' '||ch=='
'||ch=='
'||ch=='	';}
    inline int read(char *s){
        char ch=nc();
        for(;blank(ch);ch=nc());
        if(IOerror)return 0;
        for(;!blank(ch)&&!IOerror;ch=nc())*s++=ch;
        *s=0;
        return 1;
    }
    inline int RI(int &a){
        char ch=nc(); a=0;
        for(;blank(ch);ch=nc());
        if(IOerror)return 0;
        for(;!blank(ch)&&!IOerror;ch=nc())a=a*10+ch-'0';
        return 1;
    }
    struct Ostream_fwrite{
        char *buf,*p1,*pend;
        Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;}
        void out(char ch){
            if (p1==pend){
                fwrite(buf,1,BUF_SIZE,stdout);p1=buf;
            }*p1++=ch;
        }
        void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}}
        ~Ostream_fwrite(){flush();}
    }Ostream;
    inline void print(char x){Ostream.out(x);}
    inline void println(){Ostream.out('
');}
    inline void flush(){Ostream.flush();}
    char Out[OUT_SIZE],*o=Out;
    inline void print1(char c){*o++=c;}
    inline void println1(){*o++='
';}
    inline void flush1(){if (o!=Out){if (*(o-1)=='
')*--o=0;puts(Out);}}
    struct puts_write{
        ~puts_write(){flush1();}
    }_puts;
};
int n, cnt[maxn], a[maxn][11];
void GetDict() {
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < cnt[i]; ++j) {
            B[a[i][j]][i] = 1;
        }
    }
}
void work() {
    for (int i = 0; i < n; ++i) {
        fastIO::RI(cnt[i]);
        for (int j = 0; j < cnt[i]; ++j) fastIO::RI(a[i][j]);
    }
    GetDict();
    fastIO::read(s);
    int len = strlen(s);
    bitset<maxn> D;
    for (int i = 0; i < len; ++i) {
        D <<= 1; D[0] = 1;
        D &= B[s[i]-'0'];
        if (D[n-1]) {
            char temp = s[i+1]; s[i+1] = '';
            printf("%s
", s+i-n+1);
            s[i+1] = temp;
        }
    }
}
int main() {
    while (fastIO::RI(n)) work();
    return 0;
}

TLE Code kmp Ver.

#include <bits/stdc++.h>
#define maxn 1010
#define maxm 5000010
char s[maxm];
int b[maxm];
int a[maxn][11], cnt[maxn], f[maxn];
using namespace std;
typedef long long LL;
int n, m;
bool match1(int x, int y) {
    for (int i = 0; i < cnt[x]; ++i) {
        for (int j = 0; j < cnt[y]; ++j) {
            if (a[x][i] == a[y][j]) return true;
        }
    }
    return false;
}
void getfail() {
    f[0] = f[1] = 0;
    for (int i = 1; i < n; ++i) {
        int j = f[i];
        while (j && !match1(i, j)) j = f[j];
        f[i+1] = match1(i, j) ? j+1 : 0;
    }
}
bool match2(int x, int y) {
    for (int i = 0; i < cnt[y]; ++i) {
        if (b[x] == a[y][i]) return true;
    }
    return false;
}
void kmp() {
    int j = f[1];
    for (int i = 1; i < m; ++i) {
        while (j && !match2(i, j)) j = f[j];
        if (match2(i, j)) ++j;
        if (j == n) {
            for (int k = i-n+1; k <= i; ++k) putchar('0'+b[k]);
            putchar('
');
        }
    }
}
void work() {
    for (int i = 0; i < n; ++i) {
        scanf("%d", &cnt[i]);
        for (int j = 0; j < cnt[i]; ++j) scanf("%d", &a[i][j]);
    }
    getfail();
    scanf("%s", s);
    m = strlen(s);
    for (int i = 0; i < m; ++i) b[i] = s[i]-'0';
    kmp();
}
int main() {
    while (scanf("%d", &n) != EOF) work();
    return 0;
}