Number Sequence
Time Limit: 5000 MS Memory Limit: 32768 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
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
大家都在看模板，而我没有模板，iwi，一值超时，哎，然而在每次判断时先判断下首尾是否相等就可以降低时间复杂度，区别就是
Time Limit Exceed和1076MS
方法一：
#include<stdio.h>
#include<string.h>
using namespace std;
int a[1000050],b[10050];
int main()
{
int t,n,m,i,j,bj;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
scanf("%d",&m);
for(i=1; i<=n; i++)
scanf("%d",&a[i]);
for(i=0; i<m; i++)
scanf("%d",&b[i]);
for(i=1; i<=n-m+1; i++)//a数组
{
bj=1;
if(a[i]==b[0]&&a[i+m-1]==b[m-1])//判首尾
{
bj=0;
for(j=0; j<m; j++)//匹配
if(a[i+j]!=b[j])
{
bj=1;
break;
}
if(bj==0)
{
printf("%d\n",i);
break;
}
}
}
if(bj==1)
printf("-1\n");
}
}
f方法二KMP：
//引自kuangbin
#include<stdio.h>
#include<string.h>
int next[1000010];
int b[1000010];
int a[10010];
int ans;
//next[]的含义：x[i-next[i]...i-1]=x[0...next[i]-1]
//next[i]为满足x[i-z...i-1]=x[0...z-1]的最大z值（就是x的自身匹配）
void kmp_pre(int x[],int m,int next[])
{
int i,j;
j=next[0]=-1;
i=0;
while(i<m)
{
while(-1!=j&&x[i]!=x[j]) j=next[j];
next[++i]=++j;
}
}
int KMP(int x[],int m,int y[],int n)
{//x是模式串，y是主串
int i,j;
int ans=-1;
kmp_pre(x,m,next);
i=j=0;
while(i<n)
{
while(-1!=j&&y[i]!=x[j]) j=next[j];
i++;j++;
if(j>=m)
{
ans=i-m+1;break;
j=next[j];
}
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%d",&b[i]);
for(int i=0;i<m;i++)
scanf("%d",&a[i]);
int x=-1;
x=KMP(a,m,b,n);
printf("%d\n",x);
}
}
| 1076 MS |