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Number SequenceTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15007 Accepted Submission(s): 6582 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
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题解：此题就是如果匹配就输出开始匹配时的数组下标；next数组有两个含义：位置还有长度；
代码：
1 #include<stdio.h> 2 const int MAXN=10010; 3 int a[MAXN*100],b[MAXN],len1,len2,next[MAXN]; 4 void getnext(){ 5 int i=0,j=-1; 6 next[i]=j; 7 while(i<len2){ 8 if(j==-1||b[i]==b[j]){ 9 i++;j++; 10 next[i]=j; 11 } 12 else j=next[j]; 13 } 14 } 15 int kmp(){ 16 getnext(); 17 int i=0,j=0; 18 while(i<len1){ 19 if(j==-1||a[i]==b[j]){ 20 i++;j++; 21 if(j==len2)return i-j+1; 22 } 23 else j=next[j]; 24 } 25 return -1; 26 } 27 int main(){ 28 int T; 29 int N,M; 30 scanf("%d",&T); 31 while(T--){ 32 scanf("%d%d",&N,&M); 33 for(int i=0;i<N;i++)scanf("%d",&a[i]); 34 for(int i=0;i<M;i++)scanf("%d",&b[i]); 35 len1=N;len2=M; 36 printf("%d\n",kmp()); 37 } 38 return 0; 39 }
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原文地址：http://www.cnblogs.com/handsomecui/p/4711437.html
