#include <stdio.h>
#include <string.h>
int next[10005];
int str1[1000005],str2[10005];
void build_next(int len2)
{
int i=0,j=-1;
next[0] = -1;
while (i < len2)
{
if (j==-1 || str2[i] == str2[j])
{
i++;
j++;
if (str2[i] != str2[j])
{
next[i] = j;
}
else
next[i] = next[j];
}
else
j = next[j];
}
}
int KMP(int len1,int len2)
{
build_next(len2);
int i=0,j=0,cnt;
while (i < len1)
{
if (j==-1 || str1[i] == str2[j])
{
i++;
j++;
if(j==len2) //在这里判断是否为全部匹配
return i-len2+1;
}
else
j = next[j];
}
return -1; //最后还未匹配成功,直接返回-1
}
int main()
{
int N,n,m,i;
scanf("%d",&N);
while (N--)
{
scanf("%d%d",&n,&m);
for(i=0;i<n;i++)
scanf("%d",&str1[i]);
for(i=0;i<m;i++)
scanf("%d",&str2[i]);
if(n<m)
printf("-1\n");
else
{
n=KMP(n,m);
printf("%d\n",n);
}
}
return 0;
}
。。。。。。。。。。。。。。。。。。。。。。。。。。。。。
分析:题目将字符串换成了数字,原理还是一样的,简单KMP
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13443 Accepted Submission(s): 6058
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
Source
HDU 2007-Spring Programming Contest