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Number SequenceTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 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 |
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 }