Time Limit: 5000MS | Memory Limit: 32768KB | 64bit IO Format: %I64d & %I64u |
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
裸kmp,可作为模板:
#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=1000000+10;
const int maxm=10000+10;
int next[maxm],p[maxm],s[maxn];
int n,m;
void getnext(int *p)
{
int j=-1,i=0;
next[0]=-1;
while(i<m-1)
{
if(j==-1||p[i]==p[j])
{
i++,j++;
next[i]=j;
}
//else j=next[j];
else j=-1;//还是这句简洁高效
}
}
int kmp(int *p,int *s)
{
int j=0,i=0;
while(i<n&&j<m)
{
if(j==-1||s[i]==p[j])
{
i++,j++;
}
else j=next[j];
}
if(j==m) return i-j+1;
else return -1;
}
int main()
{
int i,j,t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(i=0;i<n;i++)
scanf("%d",&s[i]);
for(j=0;j<m;j++)
scanf("%d",&p[j]);
getnext(p);
printf("%d\n",kmp(p,s));
}
}
#include<cstdio>
using namespace std;
const int maxn=1000000+10;
const int maxm=10000+10;
int next[maxm],p[maxm],s[maxn];
int n,m;
void getnext(int *p)
{
int j=-1,i=0;
next[0]=-1;
while(i<m-1)
{
if(j==-1||p[i]==p[j])
{
i++,j++;
next[i]=j;
}
//else j=next[j];
else j=-1;//还是这句简洁高效
}
}
int kmp(int *p,int *s)
{
int j=0,i=0;
while(i<n&&j<m)
{
if(j==-1||s[i]==p[j])
{
i++,j++;
}
else j=next[j];
}
if(j==m) return i-j+1;
else return -1;
}
int main()
{
int i,j,t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(i=0;i<n;i++)
scanf("%d",&s[i]);
for(j=0;j<m;j++)
scanf("%d",&p[j]);
getnext(p);
printf("%d\n",kmp(p,s));
}
}
这里讲kmp原理,还不错:
http://kb.cnblogs.com/page/176818/