KMP

KMP是一种在线性时间内能处理两个字符串的包含关系的算法，例如求一个字符串里有没有另一个字符串，一个字符串里有几个另一个字符串（可重叠和不可重叠两种）。

贴个讲kmp的链接 http://blog.csdn.net/yutianzuijin/article/details/11954939

模板1(可重叠时）

#include<stdio.h>
#include<string.h>
void makeNext(const char P[],int next[])
{
    int q,k;
    int m = strlen(P);
    next[0] = 0;
    for (q = 1,k = 0; q < m; ++q)
    {
        while(k > 0 && P[q] != P[k])
            k = next[k-1];
        if (P[q] == P[k])
        {
            k++;
        }
        next[q] = k;
    }
}

int kmp(const char T[],const char P[],int next[])
{
    int n,m;
    int i,q;
    n = strlen(T);
    m = strlen(P);
    makeNext(P,next);
    for (i = 0,q = 0; i < n; ++i)
    {
        while(q > 0 && P[q] != T[i])
            q = next[q-1];
        if (P[q] == T[i])
        {
            q++;
        }
        if (q == m)
        {
            printf("Pattern occurs with shift:%d\n",(i-m+1));
        }
    }    
}

int main()
{
    int i;
    int next[20]={0};
    char T[] = "ababxbababcadfdsss";
    char P[] = "abcdabd";
    printf("%s\n",T);
    printf("%s\n",P );
    // makeNext(P,next);
    kmp(T,P,next);
    for (i = 0; i < strlen(P); ++i)
    {
        printf("%d ",next[i]);
    }
    printf("\n");

    return 0;
}

模板2(不可重叠时）

#include<stdio.h>
#include<string.h>
void makeNext(const char P[],int next[])
{
    int q,k;
    int m = strlen(P);
    next[0] = 0;
    for (q = 1,k = 0; q < m; ++q)
    {
        while(k > 0 && P[q] != P[k])
            k = next[k-1];
        if (P[q] == P[k])
        {
            k++;
        }
        next[q] = k;
    }
}

int kmp(const char T[],const char P[],int next[])
{
    int n,m;
    int i,q;
    n = strlen(T);
    m = strlen(P);
    makeNext(P,next);
    for (i = 0,q = 0; i < n; ++i)
    {
        while(q > 0 && P[q] != T[i])
            q = next[q-1];
        if (P[q] == T[i])
        {
            q++;
        }
        if (q == m)
        {
            printf("Pattern occurs with shift:%d\n",(i-m+1));
            q = 0;
        }
    }
}

int main()
{
    int i;
    int next[20]={0};
    char T[] = "ababababa";
    char P[] = "aba";
    printf("%s\n",T);
    printf("%s\n",P );
    // makeNext(P,next);
    kmp(T,P,next);
    for (i = 0; i < strlen(P); ++i)
    {
        printf("%d ",next[i]);
    }
    printf("\n");

    return 0;
}

例题

POJ 1961

题目大意，求这个字符串到i为止有多少个循环串；

int k = i-next[i];
if((i+1)%k == 0 && (i+1)!= k)
+1, (i+1)/k);

例如一个字符串的第99为指向第96位，也就是说后4-99位和前1-96位是匹配的，就是说94到96与97到99是是匹配的，而且91-94与94-96是匹配的。

一直可以推到最前面。

代码

#include <iostream>
#include <string.h>
#include <map>
#include <stdio.h>
using namespace std;
const int maxa =1000005;
int next[maxa];
int vis[maxa];
int n;
void init_kmp(char str[])
{
    memset(vis, 0, sizeof(vis));
    next[0]=-1;
    for(int i=1; str[i]!=0; i++)
    {
        int j= next[i-1];
        while(str[j+1]!=str[i]&&j>=0)
            j= next[j];
        if(str[j+1] == str[i])
            next[i] = j + 1;
        else
            next[i] = -1;
    }
}
int main()
{
    char str[maxa];
    int d =1;
    while(scanf("%d", &n), n)
    {
        scanf("%s", str);
        printf("Test case #%d\n", d++);
        init_kmp(str);
        for(int i = 0; i < n; i++)
        {
            int k = i-next[i];
            if((i+1)%k == 0 && (i+1)!= k)
                printf("%d %d\n", i+1, (i+1)/k);
        }
        printf("\n");
    }
}