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hdu 1711---KMP
题目链接
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
代码如下：
#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>using namespace std;int n[10005];int s[1000005],a[10005];int len1,len2;void kmp(){ n[0]=0; for(int i=1,k=0;i<len2;i++) { while(k>0&&a[k]!=a[i]) k=n[k-1]; if(a[k]==a[i]) k++; n[i]=k; }}int main(){ ///cout << "Hello world!" << endl; int T; cin>>T; while(T--) { scanf("%d%d",&len1,&len2); for(int i=0;i<len1;i++) scanf("%d",&s[i]); for(int i=0;i<len2;i++) scanf("%d",&a[i]); kmp(); int flag=-1; for(int i=0,j=0;i<len1;i++) { while(j>0&&a[j]!=s[i]) j=n[j-1]; if(a[j]==s[i]) j++; if(j==len2) { flag=i-len2+1; break; } } if(flag>=0) printf("%d\n",flag+1); else puts("-1"); } return 0;}
hdu 1711---KMP
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