目錄
題目描述
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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
題意分析
題意： KMP模板題，反覆研習，更得精髓。
AC代碼
#include<cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int s1[1000010],s2[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||s2[j]==s2[k])
Next[++j]=++k;
else k=Next[k];
}
}
//返回首次出現的位置
int KMP()
{
int i=0,j=0;
getNext();
while(i<n && j<m)
{
if(j==-1||s1[i]==s2[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",&s1[i]);
for(int i=0;i<m;i++)
scanf("%d",&s2[i]);
printf("%d\n",KMP());
}
return 0;
}