经典问题：字符串可分割成多少个相同重叠/不重叠子串(2087)
不可重叠：
剪花布条
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14534 Accepted Submission(s): 9196
Problem Description
一块花布条，里面有些图案，另有一块直接可用的小饰条，里面也有一些图案。对于给定的花布条和小饰条，计算一下能从花布条中尽可能剪出几块小饰条来呢？
Input
输入中含有一些数据，分别是成对出现的花布条和小饰条，其布条都是用可见ASCII字符表示的，可见的ASCII字符有多少个，布条的花纹也有多少种花样。花纹条和小饰条不会超过1000个字符长。如果遇见#字符，则不再进行工作。
Output
输出能从花纹布中剪出的最多小饰条个数，如果一块都没有，那就老老实实输出0，每个结果之间应换行。
Sample Input
abcde a3 aaaaaa aa #
Sample Output
0 3
/*------------------Header Files------------------*/
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <ctype.h>
#include <cmath>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <vector>
#include <set>
#include <limits.h>
using namespace std;
/*------------------Definitions-------------------*/
#define LL long long
#define uLL unsigned long long
#define PI acos(-1.0)
#define INF 0x3F3F3F3F
#define MOD 1000000007
#define MAX 105
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
/*---------------------Work-----------------------*/
string s,t;
int lens,lent;
int _next[1050];
void getnext()
{
int i=0,j=-1;
_next[0]=-1;
while(i<lent)
{
if(j==-1||t[i]==t[j])
{
i++,j++;
_next[i]=j;
}
else j=_next[j];
}
}
int KMP()
{
int i=0,j=0,cnt=0;
while(i<lens)
{
if(j==-1||s[i]==t[j]) i++,j++;
else j=_next[j];
if(j==lent) cnt++,j=0;
}
return cnt;
}
void work()
{
while(cin>>s)
{
if(s[0]=='#') break;
cin>>t;
lens=s.length();
lent=t.length();
getnext();
printf("%d\n",KMP());
}
}
/*------------------Main Function------------------*/
int main()
{
//freopen("test.txt","r",stdin);
work();
return 0;
}
可重叠：
Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9719 Accepted Submission(s): 3851
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
一开始和上面一样用的string，于是用cin读入，结果两次TLE，原因在于字符串长度可能很大，读入超时，然后改成字符数组+scanf读入就不超时了。
/*------------------Header Files------------------*/
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <ctype.h>
#include <cmath>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <vector>
#include <set>
#include <limits.h>
using namespace std;
/*------------------Definitions-------------------*/
#define LL long long
#define uLL unsigned long long
#define PI acos(-1.0)
#define INF 0x3F3F3F3F
#define MOD 1000000007
#define MAX 105
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
/*---------------------Work-----------------------*/
char s[1000050],t[10050];
int lens,lent;
int _next[10050];
void getnext()
{
int i=0,j=-1;
_next[0]=-1;
while(i<lent)
{
if(j==-1||t[i]==t[j])
{
i++,j++;
_next[i]=j;
}
else j=_next[j];
}
}
int KMP()
{
int i=0,j=0,cnt=0;
while(i<lens)
{
if(j==-1||s[i]==t[j]) i++,j++;
else j=_next[j];
if(j==lent) cnt++,j=_next[j];
}
return cnt;
}
void work()
{
int T; cin>>T;
while(T--)
{
getchar();
scanf("%s%s",t,s);
lens=strlen(s);
lent=strlen(t);
getnext();
printf("%d\n",KMP());
}
}
/*------------------Main Function------------------*/
int main()
{
//freopen("test.txt","r",stdin);
work();
return 0;
}