nyoj 题目5 Binary String Matching

Binary String Matching

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

 

描述

Given two strings A and B, whose alphabet consist only ‘0’ and ‘1’. Your task is only to tell how many times does A appear as a substring of B? For example, the text string B is ‘1001110110’ while the pattern string A is ‘11’, you should output 3, because the pattern A appeared at the posit

 

输入

The first line consist only one integer N, indicates N cases follows. In each case, there are two lines, the first line gives the string A, length (A) <= 10, and the second line gives the string B, length (B) <= 1000. And it is guaranteed that B is always longer than A.

输出

For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.

样例输入

3
11
1001110110
101
110010010010001
1010
110100010101011 

样例输出

3
0
3 

此题第一感觉就是用KMP算法，代码如下

#include <cstdio>
#include <cstring>

char a[12];
char b[1002];
int next[12];

void getNext() {
    next[0] = -1;
    int len = strlen(a);
    int i = 0,j = -1;
    while(i < len) {
        if(j == -1 || a[i] == a[j]) {
            i++,j++;
            next[i] = j;
        }
        else {
            j = next[j];
        }
    }
}

int calCnt() {
    int at = 0;
    int bt = 0;
    int ans = 0;
    int lena = strlen(a);
    int lenb = strlen(b);
    while(bt < lenb) {
        if(at == -1 || a[at] == b[bt]) {
            at++;
            bt++;
            if(at == lena) {
                ans++;
                at = next[lena];
            }
            continue;
        }
        if(a[at] != b[bt]) {
            at = next[at];
        }

    }
    return ans;
}

int main(int argc, char const *argv[])
{
    int n;
    while(scanf("%d",&n) != EOF) {
        while(n--) {
            scanf("%s",a);
            scanf("%s",b);
            getNext();
            int ans = calCnt();
            printf("%d
", ans);
        }
    }
    return 0;
}

此算法第一要求出next数组。而求next数组的过程本身也是一个自己和自己匹配的过程。此处用i不断前进,j不断返回，用作匹配。

计数时是新的匹配过程。和求next数组的过程神似。

若求nextval数组可能会更快些，代码如下

#include <cstdio>
#include <cstring>

char a[12];
char b[1002];
int nextval[12];

void getnextval() {
    nextval[0] = -1;
    int len = strlen(a);
    int i = 0,j = -1;
    while(i < len) {
        if(j == -1 || a[i] == a[j]) {
            i++,j++;
            if(i < len && a[i] == a[j]) {
                nextval[i] = nextval[j];
            }
            else {
                nextval[i] = j;
            }
            
        }
        else {
            j = nextval[j];
        }
    }
}

int calCnt() {
    int at = 0;
    int bt = 0;
    int ans = 0;
    int lena = strlen(a);
    int lenb = strlen(b);
    while(bt < lenb) {
        if(at == -1 || a[at] == b[bt]) {
            at++;
            bt++;
            if(at == lena) {
                ans++;
                at = nextval[lena];
            }
            continue;
        }
        if(a[at] != b[bt]) {
            at = nextval[at];
        }

    }
    return ans;
}

int main(int argc, char const *argv[])
{
    int n;
    while(scanf("%d",&n) != EOF) {
        while(n--) {
            scanf("%s",a);
            scanf("%s",b);
            getnextval();
            int ans = calCnt();
            printf("%d
", ans);
        }
    }
    return 0;
}

要注意15行的条件。但实际运行好像并没有更快。

今天偶然发现nyoj可以查看优秀的代码，这一点简直完爆其他oj，看到这样一种非常取巧的办法

#include <iostream>
#include <string>
using namespace std;

int main(int argc, char const *argv[])
{
    string s1,s2;
    int n;
    cin >> n;
    
    while(n--) {
        cin >> s1;
        cin >> s2;
        size_t  m = s2.find(s1,0);
        int ans = 0;
        while(m != string::npos) {
            ans++;
            m = s2.find(s1,m+1);
        }
        cout << ans << endl;
    }    
    
    return 0;
}