KMP算法

KMP算法

HDU 1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 57954 Accepted Submission(s): 23132

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

1
2
3
4
5
6
7
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

1
2
6
-1

Source

HDU 2007-Spring Programming Contest

代码

++
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#include<iostream>
#include<cstdio>
using namespace std;
namespace iNx{
const int N = 1e6 + 10 ;
const int M = 1e4 + 10 ;
int a[N],b[M],next[M];
int n,m;
void init(){
int i,k;
for(i=0;i<n;i++) scanf("%d",&a[i]);
for(i=0;i<m;i++) scanf("%d",&b[i]);
next[0]=-1;
i=0;
k=-1;
while(i<m-1){
if(k==-1||b[i]==b[k]){
i++;
k++;
if(b[i]!=b[k]) next[i]=k;
else next[i]=next[k];
}
else
k=next[k];
}
}
void work(){
scanf("%d%d",&n,&m);
if(n<m){
printf("-1\n");
return ;
}
init();
int i=0,j=0;
while(i<n&&j<m){
if(j==-1||a[i]==b[j]) i++,j++;
else j=next[j];
}
if(j==m)
printf("%d\n",i-j+1);
else
printf("-1\n");
}
}
int main(){
int T;
scanf("%d",&T);
while(T--) iNx::work();
return 0;
}
#

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