hdu 1686 Oulipo
Oulipo
http://acm.hdu.edu.cn/showproblem.php?pid=1686
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Problem Description
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.
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.
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.
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
题目大意:输出第一个字符串在第二个字符串中出现的次数
KMP模板题
初次写,1个错误:
getnext函数中i的枚举要从1开始,不能从0,
如果从0开始,j=f[0]=0,i=0 ,两者相等 f[1]=1
然后j=f[1]=1,i=1,两者相等
……
会导致一直相等
因为f[i]表示的是0——i-1位的最大匹配长度
#include<cstdio> #include<cstring> #include<iostream> using namespace std; int n,len1,len2,f[1000001],ans; char s1[1000001],s2[1000001]; void getnext() { for(int i=1;i<len1;i++) { int j=f[i]; while(j&&s1[i]!=s1[j]) j=f[j]; f[i+1]= s1[i]==s1[j] ? j+1 : 0; } } void getans() { int j=0; for(int i=0;i<len2;i++) { while(j&&s1[j]!=s2[i]) j=f[j]; if(s1[j]==s2[i]) j++; if(j==len1) ans++; } printf("%d\n",ans); } int main() { scanf("%d",&n); while(n--) { cin>>s1>>s2; len1=strlen(s1);len2=strlen(s2); memset(f,0,sizeof(f)); getnext(); getans(); ans=0; } }
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
//由于字符串位置从0开始存 //所以f[i]在数值上=最大前缀子串=i的最大后缀子串=下一个要检验的位置 //也就是保证了 0——f[i]-1是相等的,再匹配i与f[i] #include<cstdio> #include<cstring> #include<iostream> using namespace std; int n,len1,len2,f[1000001],ans; char s1[1000001],s2[1000001]; void getnext() { for(int i=1;i<len1;i++)//注意这里从1开始 { int j=f[i];//j=下一个要匹配的位置 while(j&&s1[i]!=s1[j]) j=f[j]; f[i+1]= s1[i]==s1[j] ? j+1 : 0;//j+1:字符串从0开始,所以长度+1 } } void getans() { int j=0; for(int i=0;i<len2;i++)//这里从1开始 { while(j&&s1[j]!=s2[i]) j=f[j]; if(s1[j]==s2[i]) j++; if(j==len1) ans++; } printf("%d\n",ans); } int main() { scanf("%d",&n); while(n--) { cin>>s1>>s2; len1=strlen(s1);len2=strlen(s2); memset(f,0,sizeof(f)); getnext(); getans(); ans=0; } }