bzoj4503: 两个串 bitset

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题目链接

bzoj4503: 两个串

题解

暴一发bitset

f[i][j] 表示 S[1..i] 是否有个后缀能匹配 T[1..j]

那么假设 S[i+1] 能匹配 T[s]，令 f[i+1][s] | = f[i][s-1]

所以预处理理出每个字符能匹配 T的哪些位置,设为[c]

那么 f[i]=((f[i-1]<<1)|(1<<1)) & mat[S[i]]

直接在mat上做匹配就好了

时间复杂度：O(|S||T|/32)

代码

#include<cstdio>

#include<bitset>

#include<cstring>

#include<algorithm>

#define LL long long

#define gc getchar()

#define pc putchar

#define LD long double

inline int read() {

    int x = 0,f = 1;

    char c = gc;

    while(c < '0' || c > '9' )c = gc;

    while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;

    return x * f ;

}

void print(int x) {

    if(x >= 10) print(x / 10);

    pc(x % 10 + '0');

}

const int maxn = 100007;

char s[maxn],t[maxn];

std::bitset<maxn>mat[30],ans;

int main() {

    scanf("%s%s",s + 1,t + 1);

    int n = strlen(s + 1),m = strlen(t + 1);

    for(int i = 1;i <= n;++ i) mat[s[i] - 'a'].set(i);

    ans.set();

    for(int i = 1;i <= m;++ i)

        if(t[i] != '?') ans &= (mat[t[i] - 'a'] >> (i - 1));

    int cnt = 0;

    for(int i = 1;i <= n - m + 1;++ i) if(ans[i] == 1) cnt ++;

    print(cnt);

    pc('\n');

    for(int i = 1;i <= n - m + 1;++ i) if(ans[i] == 1)print(i - 1),pc('\n');

    return 0;

}