hdu 1711
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Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8122 Accepted Submission(s): 3686
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
213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1
Sample Output
6-1
Source
HDU 2007-Spring Programming Contest
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lcy
代码:
#include<iostream>using namespace std;const int MAX=1000000+10;int a[MAX];int b[10000+10];int next[10000+10];void get_next(int *t,int len){ int i=0,j=-1; next[0]=-1; int Lt=len; while(i<Lt) { if(j==-1||t[i]==t[j]) { ++i,++j; next[i]=j; } else j=next[j]; }}int KMP(int a[],int b[],int lena,int lenb){ int i=0,j=0; while (i<lenb && j<lena) { if (j == -1 || a[j] == b[i])++i,++j; else j=next[j]; } if (j == lena)return i-j+1; return -1;}int main(){ int n,m,t; cin>>t; while (t--) { cin>>n>>m; for (int i=0;i<n;++i)scanf("%d",&a[i]); for (int i=0;i<m;++i)scanf("%d",&b[i]); get_next(b,m); cout<<KMP(b,a,m,n)<<endl; } return 0;}
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