hdu_1711_kmp板子

最新推荐文章于 2021-04-06 14:09:40 发布

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本文介绍了一种用于字符串匹配的高效算法——KMP算法，并通过一个具体的编程实例详细展示了该算法的工作原理及其应用过程。该算法能够有效解决在大规模数据中寻找特定子序列的问题。

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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.
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<bits/stdc++.h>
using namespace std;
const int N=(int)1e6+10;
int s[N],st[N],n,m,nxt[N];
void getnext()
{
    int i=0,j=-1;
    nxt[0]=-1;
    while(i<m)
    {
        if(j==-1||st[i]==st[j])
        {
            i++,j++;
            nxt[i]=j;
        }
        else j=nxt[j];
    }
}
int kmp(int *s,int *st)
{
    getnext();
    int i=0,j=0;
    while(i<n)
    {
        if(j==-1||s[i]==st[j])
        {
            i++,j++;
        }
        else
            j=nxt[j];
        if(j==m)
            return i-m+1;
    }
    return -1;
}
int main()
{
    int t,i,j;
    while(cin>>t)
    {
        while(t--)
        {
            scanf("%d%d",&n,&m);
            for(i=0;i<n;i++) scanf("%d",s+i);
           for(i=0;i<m;i++) scanf("%d",st+i);
           cout<<kmp(s,st)<<endl;
        }
    }
    return 0;
}