kmp入门小结

void get_next(char *s)
{int len = strlen(s);int j = 0; int k = -1;while (j < len){if (k == -1 || s[j] == s[k]){j++; k++; next[j] = k;}else k = next[k];}
}

设t = next[i]; next[i] 表示的是 i之前最大的t满足 s[0...t-1] = s[i-t...i-1]

比如 0123 4 0123 5 ，next[9] = 4.

结论：len - next[len] 为最小覆盖子串。

Poj3461 Oulipo

题意：给出两个串，问第二个串在第一个串中出现的次数

/* ***********************************************Author        : 一个西瓜Mail          : 879447570@qq.comCreated Time  : 2015-04-07 16:24:21Problem       : Oulipo
************************************************ */
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
using namespace std; 
#define INF 1000000000
//typedef __int64 LL; const int maxn = 11111;
int next[maxn];
char str[maxn];
char str1[maxn*100];void get_next(char *s)
{next[0] = -1;int j = 0 ;int k = -1;int len = strlen(s);while(j<len){if(k==-1||s[j]==s[k]){j++;k++;next[j] = k;}else k = next[k];}
}int gao(char *s1,char *s2)
{int ans = 0;int len1 = strlen(s1);int len2 =strlen(s2);int j = -1;int k = -1;while(j<len2){if(k==-1||s1[k]==s2[j]){j++;k++;//printf("%d %d\n",j,k);
        }else k = next[k];if(k==len1){ans++;k = next[k];}}return ans;
}int main()
{int T;cin>>T;while(T--){cin>>str;cin>>str1;get_next(str);int k = gao(str,str1);cout<<k<<endl;}return 0;
}

View Code

Poj1961 Period

就用到上面那个结论了

/* ***********************************************Author        : 一个西瓜Mail          : 879447570@qq.comCreated Time  : 2015-04-07 17:50:15Problem       : Period
************************************************ */
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
using namespace std; 
#define INF 1000000000
//typedef __int64 LL; const int maxn = 1111111;
int next[maxn];
char str[maxn];
void get_next(char *s)
{int len =strlen(s);int j = 0;int k = -1;next[0] = -1;while(j<len){if(k==-1||s[j]==s[k]){j++;k++; next[j] = k;}else k = next[k];}
}int main()
{int Icase = 0 ;int n;while(scanf("%d",&n)&&n){if(Icase) cout<<endl;printf("Test case #%d\n",++Icase);scanf("%s",str);get_next(str);int len = strlen(str); for(int i = 1;i<len;i++){int t = i+1;if(t%(t - next[t])==0&&(t/(t-next[t])!=1)){printf("%d %d\n",t,t/(t - next[t]));}}}return 0;
}

View Code

Poj2752 Seek the Name, Seek the Fame

给出一个串，问哪些既是前缀，又是后缀。

next数组不就是，到当前串的第i个位置，既是这个子串的前缀又是后缀的最长前缀么。迭代下去求就行了。

/* ***********************************************Author        : 一个西瓜Mail          : 879447570@qq.comCreated Time  : 2015-04-07 17:28:12Problem       : Seek the Name, Seek the Fame
************************************************ */
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
using namespace std; 
#define INF 1000000000
//typedef __int64 LL; const int maxn = 4*1e5 +10;
int next[maxn];
char str[maxn];void get_next(char *s)
{next[0] = -1;int len =strlen(s);int j = 0 ;int k = -1;while(j<len){if(k==-1||s[j]==s[k]){j++;k++;next[j] =  k;}else k = next[k];}
}
vector<int> q;
int main()
{while(scanf("%s",str)!=EOF){get_next(str);q.clear();int k = strlen(str);q.push_back(k);k = next[k];while(k){q.push_back(k);k = next[k];}for(int i = q.size() - 1;i>=0;i--){printf("%d ",q[i]);}cout<<endl;}return 0;
}

View Code

Poj2406 Power Strings

和1961一样的。

/* ***********************************************Author        : 一个西瓜Mail          : 879447570@qq.comCreated Time  : 2015-04-07 17:50:40Problem       : Power Strings
************************************************ */
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
using namespace std; 
#define INF 1000000000
//typedef __int64 LL; const int maxn = 1e6+10;
char str[maxn];
int next[maxn];void get_next(char *s)
{int len =strlen(s);int j = 0 ;int k = -1;next[0] = -1;while(j<len){if(k==-1||s[j]==s[k]){j++;k++; next[j] = k;}else k = next[k];}
}int main() {while(cin>>str){if(str[0]=='.') break;get_next(str);int len = strlen(str);if(len%(len-next[len])==0)cout<<len / (len - next[len])<<endl;else cout<<1<<endl;}return 0; 
}