杭电1711_Number Sequence
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
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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
Sample Output
6 -1
#include<stdio.h> #define Max 1000000+10 int p[10002],s[Max],next[10002]; int n,m; void getNext(int *p,int *next){ int i=0,j=-1; next[0]=-1; while(i<m){ if(j==-1 || p[i]==p[j] ) ++i, ++j, next[i]=j; else j=next[j]; } } int KMPmatch(int *s,int *p){ int i=0,j=0; getNext(p,next); while(i<n){ if(j==-1||s[i]==p[j]) i++,j++; else j=next[j]; if(j==m) return i-m+1; } return -1; } int main(){ int T,i,j; while(scanf("%d",&T)!=EOF){ while(T--){ scanf("%d%d",&n,&m); for(i=0;i<=n-1;++i) scanf("%d",&s[i]); for(i=0;i<=m-1;++i) scanf("%d",&p[i]); int sum=KMPmatch(s,p); printf("%d\n",sum); } } return 0; }
此题有固定的算法模板，记住模版就好。