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\n", 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\n", contain());
70 }
71 return 0;
72 }