KMP算法
KMP的基处题目，数字数组的KMP算法应用。
主要是next[]数组的构造，next[]存储的是字符的当前字串，与子串前字符匹配的字符数。
移动位数 = 已匹配的字符数 - 对应的部分匹配值
Problem Description
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.
Input
The 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].
Output
For 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
Source
HDU 2007-Spring Programming Contest
#include<stdio.h>
int a[],b[],next[];
void getnext(int m){
int i=,j=;
next[]=;
while(i<m){
if(j==||b[i]==b[j]){
i++; j++; next[i]=j;
}
else j=next[j];
}
} void getk(int n,int m){
int i=,j=;
while(i<=n&&j<=m){
if(j==||a[i]==b[j]){i++; j++;}
else j=next[j];
}
if(j>m) printf("%d\n",i-m);
else printf("-1\n");
} int main()
{
int t,n,m,i,j;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
for(i=;i<=n;i++) scanf("%d",&a[i]);
for(i=;i<=m;i++) scanf("%d",&b[i]);
getnext(m);
getk(n,m);
}
return ;
}