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
非常久没有做kmp的题了,这几天要做一下。这是道模板题,直接套模板即可。
#include<stdio.h> #include<string.h> int s1[1000005],s2[10006],next[10006],n,m; void nextt() { int i,j; i=0;j=-1; memset(next,-1,sizeof(next)); while(i<m){ if(j==-1 || s2[i]==s2[j]){ i++;j++;next[i]=j; } else j=next[j]; } } int kmp() { int i,j; i=0;j=0; while(i<n && j<m){ if(j==-1 || s1[i]==s2[j]){ i++;j++; } else j=next[j]; } if(j>=m){ return i+1-m; } else return 0; } int main() { int i,j,T; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); for(i=0;i<n;i++){ scanf("%d",&s1[i]); } for(i=0;i<m;i++){ scanf("%d",&s2[i]); } nextt(); if(kmp()){ printf("%d ",kmp()); } else printf("-1 "); } return 0; }