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          : [email protected]
     Created Time  : 2015-04-07 16:24:21
    Problem       : 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          : [email protected]
     Created Time  : 2015-04-07 17:50:15
    Problem       : 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          : [email protected]
     Created Time  : 2015-04-07 17:28:12
    Problem       : 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          : [email protected]
     Created Time  : 2015-04-07 17:50:40
    Problem       : 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;
}