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【KMP】Number Sequence
KMP算法
KMP的基处题目，数字数组的KMP算法应用。
主要是模式串的处理，当模式串内有重复时，模式串向左回溯重复的点的位置（next[]）。
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
1 #include<stdio.h> 2 int a[1000001],b[10001],next[10001]; 3 void getnext(int m){ 4 int i=1,j=0; 5 next[1]=0; 6 while(i<m){ 7 if(j==0||b[i]==b[j]){ 8 i++; j++; next[i]=j; 9 }10 else j=next[j];11 }12 }13 14 void getk(int n,int m){15 int i=1,j=1;16 while(i<=n&&j<=m){17 if(j==0||a[i]==b[j]){i++; j++;}18 else j=next[j];19 }20 if(j>m) printf("%d\n",i-m);21 else printf("-1\n");22 }23 24 int main()25 {26 int t,n,m,i,j;27 scanf("%d",&t);28 while(t--){29 scanf("%d%d",&n,&m);30 for(i=1;i<=n;i++) scanf("%d",&a[i]);31 for(i=1;i<=m;i++) scanf("%d",&b[i]);32 getnext(m);33 getk(n,m);34 }35 return 0;36 }
【KMP】Number Sequence
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