kuangbin专题十六 KMP&&扩展KMP HDU1711 Number Sequence

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

InputThe first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].
OutputFor each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.
Sample Input

2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output

6
-1

首先是预处理Next数组，初始化Next[0]=-1,j=-1  然后从i=0开始，如果j!=-1&&不匹配，j=Next[j]。否则判断p[++i]==p[++j],如果相等，直接让Next[i]=Next[j]，不等的话，让Next[i]=j;

然后在主串里匹配。i=0,j=0; j!=-1&&不匹配， j=Next[j]。否则i++，j++

#include<stdio.h>
#include<string.h>
int Next[10010],t[1000010],p[10010];//模式串对应的Next
int n,m,_;

void prekmp() {//预处理Next数组
    int i,j;
    j=Next[0]=-1;
    i=0;
    while(i<m) {
        while(j!=-1&&p[i]!=p[j]) j=Next[j];
        if(p[++i]==p[++j]) Next[i]=Next[j];//会快一点 博客体现了这一点
        else Next[i]=j;
    }
}

int kmp() {//查找在主串的位置
    int i=0,j=0;
    prekmp();
    while(i<n&&j<m) {
        while(j!=-1&&t[i]!=p[j]) j=Next[j];
        i++;j++;
    }
    if(j==m) return i-m+1;
    else return -1;
}

int main() {
    for(scanf("%d",&_);_;_--) {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++) {
            scanf("%d",&t[i]);
        }
        for(int i=0;i<m;i++) {
            scanf("%d",&p[i]);
        }
        int ans=kmp();
        printf("%d\n",ans);
    }
}