题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1711
Time 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
题目大意:给出两个数字串:文本串和模式串,求模式串在文本串中第一次出现的位置。
解题思路:依然是KMP模板题……
1 #include<cstdio> 2 #include<cstring> 3 #define MAXpat 10000+5 4 #define MAXstr 1000000+5 5 int len1,len2; 6 int Next[MAXpat],str[MAXstr],pat[MAXpat]; 7 void getNext() 8 { 9 int i=0, j=-1; 10 Next[0]=-1; 11 while(i<len2) 12 { 13 if(j == -1 || pat[i] == pat[j]) Next[++i]=++j; 14 else j=Next[j]; 15 } 16 } 17 int kmp() 18 { 19 getNext(); 20 int i=0, j=0; 21 while(i<len1) 22 { 23 if(j == -1 || str[i] == pat[j]) i++, j++; 24 else j=Next[j]; 25 if(j == len2) return i-j+1; 26 } 27 return -1; 28 } 29 int main() 30 { 31 int t; 32 scanf("%d",&t); 33 while(t--) 34 { 35 scanf("%d%d",&len1,&len2); 36 for(int i=0;i<len1;i++) scanf("%d",&str[i]); 37 for(int i=0;i<len2;i++) scanf("%d",&pat[i]); 38 printf("%d ",kmp()); 39 } 40 }