Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 39408 Accepted Submission(s): 16269
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
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
-1
Source
Recommend
分析：
题目意思：
给你一个长的文本串，一个短的模板串
问你模板串匹在文本串中匹配到的位置
没有匹配到的话，输出-1
跑kmp就好
#include<cstdio>
#include<iostream>
#include<cstring>
#include<memory>
using namespace std;
int moban[],wenben[];
int next1[];
int sum;
void getnext(int* s,int* next1,int m)
{
next1[]=;
next1[]=;
for(int i=;i<m;i++)
{
int j=next1[i];
while(j&&s[i]!=s[j])
j=next1[j];
if(s[i]==s[j])
next1[i+]=j+;
else
next1[i+]=;
}
}
void kmp(int* ss,int* s,int* next1,int n,int m)
{
int ans=-;
getnext(s,next1,m);
int j=;
for(int i=;i<n;i++)
{
while(j&&s[j]!=ss[i])
j=next1[j];
if(s[j]==ss[i])
j++;
if(j==m)
{
ans=i-m+;
break;
}
}
printf("%d\n",ans);
}
int main()
{
int t;
scanf("%d",&t);
int n,m;
while(t--)
{
scanf("%d %d",&n,&m);
for(int i=;i<n;i++)
scanf("%d",&wenben[i]);
for(int i=;i<m;i++)
scanf("%d",&moban[i]);
kmp(wenben,moban,next1,n,m);
}
return ;
}