NYOJ 5 Binary String Matching (kmp 字串匹配)
阿新 • • 發佈:2018-12-24
Binary String Matching
時間限制:3000 ms | 記憶體限制:65535 KB 難度: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
- 輸入
- 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.
- 輸出
- For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.
- 樣例輸入
-
3 11 1001110110 101 110010010010001 1010 110100010101011
- 樣例輸出
-
3 0 3
- 來源
- 網路
- 上傳者
- 南陽oj上這道題太水了;直接用樸素演算法暴力就水過了,資料量太少了,然後我們把題目移到我們自己的oj上,自己多出了一些資料,測試了一下,樸素演算法用時3300ms,kmp用時1700ms左右,僅供參考,足以顯示出kmp演算法的優越性啊,最近兩天也看了好久kmp,感覺有些細節的地方還是理解的不到位;還是要多回顧理解,看的是july大神的部落格kmp演算法 講的比較詳細,多讀幾遍,大話資料結構上分析的也比較好;供自己以後複習回顧;
- 下面是樸素演算法過的,暴力;
- 樸素演算法關鍵在於回溯,處理好回溯;
下面是kmp演算法,還要多多加強,鞏固理解;#include <cstdio> #include <cstring> int main() { int n,count; char a[200],b[1200]; scanf("%d",&n); getchar(); while(n--) { count=0; int i=0,j=0,len; scanf("%s\n%s",a,b); len=strlen(b); while(i<=len) { if(a[j]=='\0') { count++; i=i-j+1; j=0; } else if(a[j]==b[i]) { i++; j++; } else { i=i-j+1; //關鍵在於回溯 j=0; } } printf("%d\n",count); } return 0; }
- 樸素演算法中的回溯很多是不必要的,所以就有了kmp演算法,用一個next陣列儲存下一次要匹配的位置,這裡難理解的就是next陣列,其中還要理解字尾和字首的關係,通過匹配字串前後的聯絡,得出next陣列;而就不需要移動主串的位置;節約了回溯的時間;
#include<cstdio> #include<cstring> int nextval[200]; void get_next(char a[])//得到next陣列; { int len; int i=0,j=-1; nextval[0]=-1; len=strlen(a); while(i<=len) { if(j==-1 || a[i]==a[j]) { ++i; ++j; if(a[i]==a[j]) nextval[i] = nextval[j]; //把回溯的內容全換成是next陣列; else nextval[i] = j; } else j=nextval[j]; } } int kmp(char a[],char b[])//kmp的主體函式 { int i=0,j=0,count=0; int lena,lenb; lena=strlen(a); lenb=strlen(b); get_next(a); while(i<=lenb) { if(j==-1 || a[j]==b[i]) { ++i; ++j; } else j=nextval[j]; if(j>=lena) { count++; j=nextval[j]; } } return count; } int main() { int n; char a[20],b[1200]; scanf("%d",&n); while(n--) { scanf("%s\n%s",a,b); printf("%d\n",kmp(a,b)); } return 0; }