HDU 1711 Number Sequence （kmp 已被搞死。）

时间：2015-03-26 09:12:30 收藏：0 阅读：233

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12603 Accepted Submission(s): 5735

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

Source

HDU 2007-Spring Programming Contest

给你两个窜、

求B窜匹配A窜时第一个数字在A窜里的下标。。。

KMP算法。搞了半天求Next【】数组的还不是很懂。。

先上代码。。。

以后再写myself的想法。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <cmath>
#include <cstring>
#include <queue>
#include <stack>
#include <time.h>
using namespace std;
typedef long long ll;
#define mod 1e9+9
#define inf 0x7f7f7f7f
int a[1000005];
int b[10005];
int Next[10005];
int n,m;
void GetNext()   //这个GetNext研究了一个晚上还是不太懂。。
{
    int i,j;
    i=0;
    j=-1;
    Next[0]=-1;
    while(i<m)
    {
        if(j==-1 ||b[i]==b[j])
        {
            ++i;
            ++j;
            if(b[i]!=b[j])
                Next[i]=j;
            else
                Next[i]=Next[j];

        }
        else
            j=Next[j];
    }
}
int kmp()
{
    int i,j;
    i=j=0;
    while(i<n)
    {
        if(j==-1 ||a[i]==b[j])
        {
            ++i;
            ++j;
            if(j==m)
                return i-m+1;
        }
        else
            j=Next[j];
    }
    return -1;
}
int main()
{
      int i,t;
      cin>>t;
      while(t--)
      {
          cin>>n>>m;
          for(i=0;i<n;i++)
            cin>>a[i];
          for(i=0;i<m;i++)
            cin>>b[i];
          GetNext();
          cout<<kmp()<<endl;
      }
      return 0;
}

原文：http://blog.csdn.net/sky_miange/article/details/44629095