# kmp算法专题

kmp算法前置技能：无

kmp算法是一种高效的字符串匹配算法，对于在给定长为n的主字符串S里查找长为m的模式字符串P，可以将时间复杂度从O(n*m)优化为O(n+m)。

kmp算法的核心是一个被称为部分匹配表(Partial Match Table)(下文简称为PMT)的数组。对于一个字符串“abababca”来说，它的PMT如下图的value所示，PMT的值是字符串的前缀集合与后缀集合的交集中最长元素的长度。

HDU1711 Number Sequence

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

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

int n, m;
int a[1000005], b[10005], next[10005];

void buildnext()
{
next[0] = -1;
int i = 0, j = -1;
while (i < m)
{
if (j == -1 || b[i] == b[j])
next[++i] = ++j;
else
j = next[j];
}
}

int kmp()
{
buildnext();
int i = 0, j = 0;
while (i < n && j < m)
{
if (j == -1 || a[i] == b[j])
{
++i;
++j;
}
else
j = next[j];
}
if (j == m)
return i - j;
else
return -1;
}

int main()
{
int t, i, ret;
scanf("%d", &t);
while (t--)
{
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
memset(next, 0, sizeof(next));
scanf("%d %d", &n, &m);
for (i = 0; i < n; ++i)
scanf("%d", &a[i]);
for (i = 0; i < m; ++i)
scanf("%d", &b[i]);
ret = kmp();
if (ret >= 0)
printf("%d\n", ret + 1);
else
printf("-1\n");
}
return 0;
}

posted @ 2019-06-30 14:49  redleaves  阅读(248)  评论(0编辑  收藏
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