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Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22739 Accepted Submission(s): 9727
1 //2016.10.7 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 6 using namespace std; 7 8 const int inf = 0x3f3f3f3f; 9 int a[1000005], b[10005], nex[10005]; 10 11 int kmp(int a[], int b[]) 12 { 13 int ans = 0; 14 nex[0] = -1; 15 for(int i = 0, fail = -1; b[i] != inf;)//求nex数组, fail为失配指针 16 { 17 if(fail==-1 || b[i] == b[fail]) 18 { 19 i++, fail++; 20 nex[i] = fail; 21 }else fail = nex[fail]; 22 } 23 int i = 0, j = 0; 24 for(; a[i] != inf; i++, j++) 25 { 26 if(j != -1 && b[j] == inf)return i-j+1; 27 while(j != -1 && a[i] != b[j])j = nex[j]; 28 } 29 if(b[j] == inf)return i-j+1; 30 return -1; 31 } 32 33 int main() 34 { 35 int T, n, m; 36 scanf("%d", &T); 37 while(T--) 38 { 39 scanf("%d%d", &n, &m); 40 for(int i = 0; i < n; i++)scanf("%d", &a[i]); 41 for(int i = 0; i < m; i++)scanf("%d", &b[i]); 42 a[n] = b[m] = inf;//设置截止符号 43 printf("%d\n", kmp(a, b)); 44 } 45 46 return 0; 47 }
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原文地址:http://www.cnblogs.com/Penn000/p/5936700.html