【字符串入門專題1】 I - Number Sequence hdu1711【kmp】




Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28715    Accepted Submission(s): 12063


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
  題意:找到s2在s1中第一次出現的位置,沒有找到輸出-1

思路:kmp稍微變形一下,依然可以用來處理數字字符串,只要找到s1中出現s2就直接返回下標,若在循環內沒找到,在循環結束后判斷s2是不是s1的后綴就行了


#include<stdio.h>
#define inf 99999999
#define N 1100000
#define M 11000
int a[N],b[M];
int next[M];

int kmp(int a[],int b[],int next[])//找到s2第一次在s1中出現的位置
{
int j,k,i;
j = 0;
k = -1;
next[0] = -1;
while(b[j]!=inf)//建立next數組
{
while(k!=-1&&b[k]!=b[j])
{
k = next[k];
}

j++;
k++;
if(b[k]!=b[j])
next[j] = k;
else
next[j] = next[k];
}
i = j = 0;
while(a[i]!=inf)
{
if(b[j] == inf)
return i-j+1;//第一次出現就返回在s1中的下標
else
{
while(j != -1&&a[i]!=b[j])
j = next[j];
i++;
j++;
}
}
if(b[j] == inf)//判斷s2是否為s1數組的后綴
return i-j+1;
return -1;//不存在返回-1
}

int main()
{
int t,i,j,n,m;
scanf("%d",&t);
while(t --)
{
scanf("%d%d",&n,&m);
for(i = 0; i < n; i ++)
scanf("%d",&a[i]);
for(j = 0; j < m; j++)
scanf("%d",&b[j]);
a[i] = b[j] = inf;//設置截止符

printf("%d\n",kmp(a,b,next));
}
return 0;
}



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