杭电 1711 Number Sequence（kmp）

[align=left]Problem Description[/align]
Given two sequences of numbers : a[1], a[2], ...... , a
, 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.

[align=left]Input[/align]
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
. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

[align=left]Output[/align]
For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

6
-1   [code]#include<stdio.h>
int a[1000002];
int s[10002],next[10002];
int m,n;
void get_next()//首先我们要生成一个next数组
{
int i=0,j=-1;
next[0]=-1;
while(i<m-1){
if(j==-1||s[i]==s[j])//什么？不会算next数组里面的值？赶快回炉深造去
{
i++;
j++;
next[i]=j;//其实就是一位一位的递推 多模拟几遍就会了
}
else
j=next[j];
}
}
int kmp()//开始kmp算法 主要就是两个字符串比对
{
int i=0,j=0;
get_next();
while(i<=n-1)
{
if(j==-1||a[i]==s[j])
{
i++;
j++;
}
else
j=next[j];
if(j==m)//如果j的值和m相等说明已经匹配到和s相同的一段了 我们也不需要再继续匹配了
return i-m+1;//i的值是匹配到s串的最后一位的值
}
return -1;
}
int main()
{
int test,i;
scanf("%d",&test);
while(test--)
{
scanf("%d %d",&n,&m);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<m;i++)
scanf("%d",&s[i]);
printf("%d\n",kmp());
}
return 0;
}

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