题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1711
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15683 Accepted Submission(s): 6898
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
初学KMP，水一发模版题
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <algorithm> 5 #include <iostream> 6 #include <cmath> 7 #include <queue> 8 #include <map> 9 #include <stack> 10 #include <list> 11 #include <vector> 12 13 using namespace std; 14 const int maxn = 1000010; 15 int na, nb; 16 int a[maxn]; 17 int b[maxn]; 18 int pre[maxn]; 19 20 void getpre(int *b, int *pre) { 21 int j, k; 22 pre[0] = -1; 23 j = 0; 24 k = -1; 25 while(j < nb - 1) { 26 if(k == -1 || b[j] == b[k]) {//匹配 27 j++; 28 k++; 29 pre[j] = k; 30 } 31 else { //b[j] != b[k] 32 k = pre[k]; 33 } 34 } 35 } 36 37 int kmp() { 38 int i = 0; 39 int j = 0; 40 getpre(b, pre); 41 while(i < na) { 42 if(j == -1 || a[i] == b[j]) { 43 i++; 44 j++; 45 } 46 else { 47 j = pre[j]; 48 } 49 if(j == nb) { 50 return i - nb + 1; 51 } 52 } 53 return -1; 54 } 55 56 int main() { 57 int T; 58 scanf("%d", &T); 59 while(T--) { 60 scanf("%d %d", &na, &nb); 61 for(int i = 0; i < na; i++) { 62 scanf("%d", &a[i]); 63 } 64 for(int i = 0; i < nb; i++) { 65 scanf("%d", &b[i]); 66 } 67 printf("%d ", kmp()); 68 } 69 return 0; 70 }
本题还可以用hash做，效率与kmp差距不大。
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <algorithm> 5 #include <iostream> 6 #include <cmath> 7 #include <queue> 8 #include <map> 9 #include <set> 10 #include <stack> 11 #include <list> 12 #include <vector> 13 14 using namespace std; 15 16 typedef unsigned long long ull; 17 const int B = 100007; 18 const int maxn = 1000010; 19 int a[maxn], b[maxn]; 20 int na, nb; 21 22 ull quickmul(int x, int n) { 23 ull ans = 1; 24 ull t = x; 25 while(n) { 26 if(n & 1) { 27 ans = (ans * t); 28 } 29 t = t * t; 30 n >>= 1; 31 } 32 return ans; 33 } 34 35 int contain() { 36 if(na > nb) { 37 return false; 38 } 39 ull t = quickmul(B, na); 40 ull ah = 0, bh = 0; 41 for(int i = 0; i < na; i++) { 42 ah = ah * B + a[i]; 43 } 44 for(int i = 0; i < na; i++) { 45 bh = bh * B + b[i]; 46 } 47 for(int i = 0; i + na <= nb; i++) { 48 if(ah == bh) { 49 return i + 1; 50 } 51 if(i + na < nb) { 52 bh = bh * B + b[i+na] - b[i] * t; 53 } 54 } 55 return -1; 56 } 57 int main() { 58 // freopen("in", "r", stdin); 59 int T; 60 scanf("%d", &T); 61 while(T--) { 62 scanf("%d %d", &nb, &na); 63 for(int i = 0; i < nb; i++) { 64 scanf("%d", &b[i]); 65 } 66 for(int i = 0; i < na; i++) { 67 scanf("%d", &a[i]); 68 } 69 printf("%d ", contain()); 70 } 71 return 0; 72 }