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[];
 char b[];
 int next[];
 
 void getNext() {
     next[] = -;
     int len = strlen(a);
     int i = ,j = -;
     while(i < len) {
         if(j == - || a[i] == a[j]) {
             i++,j++;
             next[i] = j;
         }
         else {
             j = next[j];
         }
     }
 }
 
 int calCnt() {
     int at = ;
     int bt = ;
     int ans = ;
     int lena = strlen(a);
     int lenb = strlen(b);
     while(bt < lenb) {
         if(at == - || 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\n", ans);
         }
     }
     return ;
 }

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[];
 char b[];
 int nextval[];
 
 void getnextval() {
     nextval[] = -;
     int len = strlen(a);
     int i = ,j = -;
     while(i < len) {
         if(j == - || 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 = ;
     int bt = ;
     int ans = ;
     int lena = strlen(a);
     int lenb = strlen(b);
     while(bt < lenb) {
         if(at == - || 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\n", ans);
         }
     }
     return ;
 }

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,);
         int ans = ;
         while(m != string::npos) {
             ans++;
             m = s2.find(s1,m+);
         }
         cout << ans << endl;
     }    
 
     return ;
 }