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; }
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; }
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; }
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; }
