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
这题是一个最简单的kmp的题目, 子序列的匹配~·
#include<stdio.h> int bbs,n,m; int s[1000010]; int a[10010]; int next[10010]; void tonext() //得到next数组 { int i=0,j=-1; next[0]=-1; while(i<m) { if(j==-1||a[i]==a[j]) { i++;j++; if(a[i]==a[j]) //如果还是相等 就继续优化 { next[i]=next[j]; } else next[i]=j; } else j=next[j]; } } int kmp() //序列的匹配 { int i=0,j=0; //2个都设为0 while(i<n&&j<m) { if(a[j]==s[i]||j==-1) //相等 就比较下一个 { i++;j++; } else j=next[j]; } if(j>m-1) //返回第几个开始 return i-m+1; else return -1; } int main() { scanf("%d",&bbs); while(bbs--) { scanf("%d%d",&n,&m); if(n<m) { printf("-1\n"); continue; } for(int i=0;i<n;i++) { scanf("%d",&s[i]); } for(int i=0;i<m;i++) { scanf("%d",&a[i]); } tonext(); int ans=kmp(); printf("%d\n",ans); } return 0; }