HDU1711 Number Sequence
HDU1711 Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3852 Accepted Submission(s): 1781
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
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
*******************************************************
题目大意：其实不必解释。
解题思路：最裸的KMP。
#include <stdio.h>
#include <string.h>
#define N 1000005
#define M 10005
int a[N],b[M];
int n,m;
int p[M];
void re(void)
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<m;i++)
scanf("%d",&b[i]);
}
void run(void)
{
memset(p,-1,sizeof(p));
for(int i=1;i<m;i++)
{
int k=p[i-1];
while(1)
{
if(b[i]==b[k+1])
{
p[i]=k+1;
break;
}
if(k==-1)break;
k=p[k];
}
}
int j=-1;
for(int i=0;i<n;i++)
{
while(1)
{
if(a[i]==b[j+1])
{
j++;
if(j+1==m)
{
printf("%d\n",i-j+1);
return ;
}
break;
}
if(j==-1)break;
j=p[j];
}
}
puts("-1");
}
int main()
{
freopen("/home/fatedayt/in","r",stdin);
int ncase;
scanf("%d",&ncase);
while(ncase--)
{
re();
run();
}
return 0;
}