http://poj.org/problem?id=3461 Oulipo Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 41051   Accepted: 16547 Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a mem…

题意:找出模板在文本串中出现的次数 思路:KMP模板题 #include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<iostream> #include<algorithm> #include<vector> #include<map> #include<queue> #include<stack&…

http://poj.org/problem?id=3461 直接KMP就好.水题 #include<cstdio> #include<cstring> const int MAXN=10000+10; const int MAXM=1000000+10; char P[MAXN],T[MAXM]; int f[MAXN],n,m,ans; void getFail() { f[0]=f[1]=0; for(int i=1;i<n;i++){ int j = f[i]; wh…

题意: 给两组字符串a和b,求a在b中出现的次数 关于KMP: 马拉车算法是处理回文串,而KMP是处理前后缀的相同字符串的最长长度. a | a | b | a | a | f | a | a 数组 p [ ] 下标为 i a | a | b | a | a | c 数组 t [ ]下标为 j -1| 0 | 1 | 0 | 1 | 2 next [ ] shu数组的值 比较这两组字符串,当p[5]!=t[5]时,如果暴力,那么 i 回溯到1,j 回溯到0,重新比较,可是在之前比较过的0~4下标…

题目:POJ3080 http://poj.org/problem?id=3080 题意:对于输入的文本串,输出最长的公共子串,如果长度相同,输出字典序最小的. 这题数据量很小,用暴力也是16ms,用后缀数组可以到0ms,但我不会XD. 暴力: #include<cstdio> #include<cstring> using namespace std; ][],ans[]; int main(){ int T,n; scanf("%d", &T); w…

Oulipo Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 23667   Accepted: 9492 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…

题目链接: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…

题意:统计其中一个子串的出现次数 题解:即KMP算法中j==m的次数 //作者:1085422276 #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include &l…

# include <stdio.h> # include <algorithm> # include <string.h> using namespace std; char a1[1000010],a2[1000010]; int next[1000010]; int len1,len2,cot; void Getnext() { int i=0,j=-1; next[0]=-1; while(i<=len1) { if(j==-1||a1[i]==a1[j]…

本题就是给出非常多对字符串,然后问一个字符串在另外一个字符串出现的次数. 就是所谓的Strstr函数啦. Leetcode有这道差点儿一模一样的题目. 使用KMP算法加速.算法高手必会的算法了. 另外看见讨论说什么使用KMP还超时,最大可能是没有真正理解next table的含义,写了错误的代码,故此尽管自己执行结果正确,可是却没有真正发挥next table的作用.使得算法退化为暴力法了,所以执行正确,但超时. KMP參考: http://blog.csdn.net/kenden23/arti…