标签：ase code ons com kmp each void 一个 find
InputThe 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 a11, a22, ...... , aNN. The third line contains M integers which indicate b11, b22, ...... , bMM. All integers are in the range of ?1000000,1000000?1000000,1000000.
OutputFor 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
题目意思:在一个长的字符串中找到一串短的字符串
解法:用KMP,有个大牛讲的很好:http://www.ruanyifeng.com/blog/2013/05/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm.html
1 #include <iostream>
2 #include <string.h>
3
4 using namespace std;
5
6 const int MAX = 1000000 + 50;
7 int a[MAX],b[MAX];
8 int visit[MAX];
9 int an,bm;
10
11 void DP_2()
12 {
13 int ai = 0;
14 int bi = 0;
15 while(ai < an )
16 {
17 // cout<<ai<<" "<<bi<<endl;
18 //cout<<a[ai]<<" "<<b[bi]<<endl;
19 if(bi==-1||a[ai] == b[bi])
20 {
21 ai++;
22 bi++;
23 }
24 else
25 {
26 bi = visit[bi];
27 }
28 if(bi==bm)
29 {
30 cout<<ai - bm+1<<endl;
31 return;
32 }
33
34 }
35 cout<<-1<<endl;
36
37
38
39 }
40
41 void DP()
42 {
43 int j = 0;
44 int k = -1;
45 visit[0] = -1;
46 while(j < bm)
47 {
48 if(k==-1||b[j]==b[k] )
49 {
50 k++;
51 j++;
52 visit[j] = k;
53 }
54 else
55 k = visit[k];
56 }
57
58 }
59
60 int main()
61 {
62 int N;
63 cin>>N;
64 while(N--)
65 {
66 memset(visit,0,sizeof(visit));
67 memset(a,0,sizeof(a));
68 memset(b,0,sizeof(b));
69
70 cin>>an>>bm;
71 for(int i = 0;i <an;i++)
72 cin>>a[i];
73 for(int i = 0;i <bm;i++)
74 cin>>b[i];
75
76 DP();
77
78 DP_2();
79
80 }
81
82 return 0;
83 }
标签：ase code ons com kmp each void 一个 find
原文地址：http://www.cnblogs.com/a2985812043/p/7270229.html
