題意：字符串匹配：尋找字符串S中，字符串T出現的次數

思路：KMP或哈希

The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:?

Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…?

Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.?

So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.?
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Input

The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:?

One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).?
One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.

Output

For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.?
?

Sample Input

3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN

Sample Output

1 3 0

代碼：KMP

/*next數組的含義就是一個固定字符串的最長前綴和最長后綴相同的長度。 ABCEFGHRHHABCABCEFGHRHHABC 末尾（ABC）能跟開頭（ABC）重合的最大長度就是前綴和子綴的最大公共長度*/ #include<stdio.h> #include<string.h> using namespace std; int n,a,b,dp[10010]; char u[10010],v[1000010]; void ne()//**注意最長前綴：是說以第一個字符開始，但是不包含最后一個字符。*/ {int i=0,j=-1;dp[0]=-1;//next[0]初始化為-1，-1表示不存在相同的最大前綴和最大后綴while(i<a){if(j==-1||u[i]==u[j]) dp[++i]=++j;//這個是把算的k的值（就是相同的最大前綴和最大后綴長）賦給next[q]else j=dp[j];//如果下一個不同，那么k就變成next[k]，注意next[k]是小于k的，無論k取任何值。}//往前回溯 } void KMP() {int ans=0;int i=0,j=0;while(i<a&&j<b){if(i==-1||u[i]==v[j])j++,i++;else i=dp[i];if(i==a){ans++;i=dp[i];}}printf("%d\n",ans); } int main() {scanf("%d",&n);while(n--){scanf("%s%s",u,v);a=strlen(u);b=strlen(v);ne();KMP();}return 0; }

哈希

#include<stdio.h>/*哈希*/ #include<string.h> #include<algorithm> using namespace std; typedef unsigned long long ull; const int M=1e4+10; const int p=13331;/*哈希的基數*/ int t,a,b; char u[M],v[100*M]; ull dp[M]; ull h[M*100]; ull k; void init() {dp[0]=1;for(int i=1;i<M;i++)dp[i]=dp[i-1]*p; } ull geth(char *s) {ull h=0;int len=strlen(s);for(int i=0;i<len;i++)h=h*p+s[i];return h; } void gethash(char *s) {int len=strlen(s);h[0]=0;for(int i=0;i<len;i++)h[i+1]=h[i]*p+s[i]; }int main() {scanf("%d",&t);init();while(t--){scanf("%s%s",u+1,v+1);k=geth(u+1);gethash(v+1);a=strlen(u+1);b=strlen(v+1);int ans=0;for(int i=0;i+a<=b;i++){ull tmp=h[i+a]-h[i]*dp[a];if(tmp==k)ans++;}printf("%d\n",ans);}return 0; }

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總結

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