KMP算法总结

KMP题目重在理解next数组的含义：

next数组的作用 ：next[j]  记录模式串中第 j 个字符的最长公共前缀长度（重要，这是它的意义所在）

                                    第二种理解方式，当模式串与主串失配时，跳回的位置。

next[len] (即字符串'\0'结束标志的next值)：单个字符串匹配时与周期有关

HDU 1711(模板题)  3336  1358  2203  2594  3746

HDU 1711

#include <string.h>  
#include <iostream>  

using namespace std;  

int a[1000005],b[10005];  
int p[10005];  
int n,m;  
void getNext () {  
    int i,j = 0;  
    p[1] = 0;  
    for (i = 2; i <= m; i++) {  
        while (j > 0 && b[j+1] != b[i]) {  
            j = p[j];  
        }  
        if (b[i] == b[j+1]) j++;  
        p[i] = j;  
    }  
}  
void kmp () {  
    int i,j = 0;  
    for (i = 1; i <= n; i++) {  
        while (j > 0 && b[j+1] != a[i]) {  
            j = p[j];  
        }  
        if (b[j+1] == a[i]) j++;  
        if (j == m) {  
            printf("%d\n",i - m +1 );  
            return ;  
        }  
    }  
    printf("%d\n",-1);  
}  
int main () {  
    int t;  
    scanf("%d",&t);  
    while (t--) {  
        scanf("%d%d",&n,&m);  
        int i,j;  
        for (i = 1; i <= n ;i++) scanf("%d",&a[i]);  
        for (i = 1; i <= m; i++) scanf("%d",&b[i]);  
        getNext();  
        kmp();  
    }  
    return 0;  
}

HDU 3746

#include <stdio.h>
#include <string.h>

#define LEN 1000001

int next[LEN];
int len;
char str[LEN];

void getNext() {
    int i, j;
    next[0] = -1;
    i = 0;
    j = -1;
    while(str[i] != '\0') {
        if(-1 == j || str[i] == str[j]) {
            ++i;
            ++j;
            next[i] = j;
        }
        else {
            j = next[j];
        }
    }
}

int main() {
    int files;
    scanf("%d", &files);
    while(files--) {
        int period;
        scanf("%s", str);
        memset(next, 0, sizeof(next));
        getNext();
        len = strlen(str);
        period = len - next[len];
        if(next[len] == 0) {
            printf("%d\n", len);
        }
        else if(len%period == 0) {
            printf("0\n");
        }
        else {
            int ans = period - (len%period);
            printf("%d\n", ans);
        }
    }
    return 0;
}