KMP / HDU 1711 找到匹配的位置并返回   2013-01-17


kmp hdu 1711 number sequence

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

Code

#include <iostream>  
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <bitset>
#include <list>
#include <map>
#include <set>
#include <iterator>
#include <algorithm>
#include <functional>
#include <utility>
#include <sstream>
#include <climits>
#include <cassert>
#define BUG puts("here!!!");

using namespace std;
const int N = 1000005;
const int M = 10005;

int s[N];
int t[M];
int next[M];
void getNext(int len) {
int i, j;
i = 0, j = -1;
next[0] = -1;
while(i < len-1) {
if(j == -1 || t[i] == t[j]) {
i++, j++, next[i] = j;
}
else {
j = next[j];
}
}
}
int kmp(int sl, int tl) {
int i = 0, j = 0;
while(i < sl && j < tl) {
if(j == -1 || s[i] == t[j]) {
i++, j++;
}
else j = next[j];
}
if(j == tl) return i-j+1;
else return -1;
}
// abcabcababcabcabdef
// abcabcabd
int main() {
int T, n, m, ans;
cin >> T;
while(T--) {
cin >> n >> m;
for(int i = 0; i < n; i++) {
scanf("%d", s+i);
}
for(int i = 0; i < m; i++) {
scanf("%d", t+i);
}
getNext(m);
ans = kmp(n, m);
cout << ans << endl;
}
return 0;
}

分享到:


  如果您觉得这篇文章对您的学习很有帮助, 请您也分享它, 让它能再次帮助到更多的需要学习的人. 您的支持将鼓励我继续创作 !
本文基于署名4.0国际许可协议发布,转载请保留本文署名和文章链接。 如您有任何授权方面的协商,请邮件联系我。

Contents

  1. Description
  2. Input
  3. Output
  4. Sample Input
  5. Sample Output
  6. Code