HDU 1711 Number Sequence (KMP找子串第一次出现的位置）

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5898 Accepted Submission(s): 2652


[align=left]Problem Description[/align]
Given two sequences of numbers : a[1], a[2], ...... , a
, 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.

[align=left]Input[/align]
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
. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

[align=left]Output[/align]
For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

6 -1

[align=left]Source[/align]
HDU 2007-Spring Programming Contest

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#include<stdio.h>
int a[1000010],b[10010];
int next[10010];
int n,m;
void getNext()
{
int j,k;
j=0;
k=-1;
next[0]=-1;
while(j<m)
{
if(k==-1||b[j]==b[k])
next[++j]=++k;
else k=next[k];
}
}
//返回首次出现的位置
int KMP_Index()
{
int i=0,j=0;
getNext();

while(i<n && j<m)
{
if(j==-1||a[i]==b[j])
{
i++;
j++;
}
else j=next[j];

}
if(j==m) return i-m+1;
else return -1;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<m;i++)
scanf("%d",&b[i]);
printf("%d\n",KMP_Index());
}
return 0;
}

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAXN=1000010;

int a[MAXN];
int b[MAXN];

int n,m;
int next[MAXN];
/*
根据定义next[0]=-1，假设next[j]=k, 即P[0...k-1]==P[j-k,j-1]
1)若P[j]==P[k]，则有P[0..k]==P[j-k+1,j]，很显然，next[j+1]=next[j]+1=k+1;
2)若P[j]!=P[k]，则可以把其看做模式匹配的问题，即匹配失败的时候，k值如何移动，显然k=next[k]。
*/

void getNext()
{
int j,k;
next[0]=-1;
j=0;
k=-1;
while(j<m)
{
if(k==-1||b[j]==b[k])
{
j++;
k++;
next[j]=k;
}
else k=next[k];
}
}

int KMP_Index()
{
int i=0,j=0;
getNext();
while(i<n&&j<m)
{
if(j==-1||a[i]==b[j])
{
i++;j++;
}
else j=next[j];
}
if(j==m)return i-m+1;
else return -1;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)scanf("%d",&a[i]);
for(int i=0;i<m;i++)scanf("%d",&b[i]);
printf("%d\n",KMP_Index());
}
return 0;
}