Number Sequence kmp吧模板题

最新推荐文章于 2021-08-26 12:17:14 发布

原创最新推荐文章于 2021-08-26 12:17:14 发布 · 125 阅读

CC 4.0 BY-SA版权

本文深入探讨了KMP算法的原理与实现，通过具体案例展示了如何使用KMP算法解决字符串匹配问题。文章首先介绍了KMP算法的基本概念，然后详细解释了Next数组的设置过程，最后通过一个实际的例子说明了KMP算法的具体应用。适合对字符串匹配算法感兴趣或需要解决相关问题的读者。

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

kmp模板题熟能生巧

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mc=1e4+5;
const int maxn=1e6+5;
int str[maxn],pattern[mc];
int PI[mc];
void cpf(int *p,int m)  //设置Next数组
{
     PI[0]=-1,PI[1]=0;
    for(int i=1;i<m;i++)
    {
        int j=PI[i];
        while(j>0&&p[i]!=p[j]) j=PI[j];
        if(p[j]==p[i])
            PI[i+1]=j+1;
        else PI[i+1]=0;

	}
   // printf("%d ",PI[0]);
}
void kmp(int *s,int n,int *p,int m)
{
    int num=0;
    int j=0;
    cpf(p,m);
    for(int i=0;i<n;i++)
    {

        while(j>0&&s[i]!=p[j])
		 j=PI[j];
        if(s[i]==p[j])
            j++;
        if(j==m)
        {
            num=i+2-m;
            break;
        }
    }
    if(num==0)
        printf("-1\n");
    else
    printf("%d\n",num);
}
int main()
{

    int Test;
    scanf("%d",&Test);
    while(Test--)
    {
    	memset(PI,0,sizeof(PI));
        int n,m;
        if(n<m)
        continue;
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%d",&str[i]);
        for(int i=0;i<m;i++)
            scanf("%d",&pattern[i]);
            kmp(str,n,pattern,m);
    }
    return 0;
}