HDU - 1711 Number Sequence(KMP入門模板題)


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:輸入測試次數t,輸入主串長度n和模式串長度m,輸出模式串在主串中第一次出現的位置,若主串中不含模式串,輸出-1。
裸KMP模板題,注意KMP函數最后是匹配完模式串時指針在主串中的位置。還需減掉模式串長度才是第一次出現的位置。

代碼如下:

#include<stdio.h>///裸KMP模板
void kmp_pre(int x[],int m,int next[])
{
    int i,j;
    j=next[0]=-1;
    i=0;
    while(i<m)
    {
        while(j!=-1&&x[i]!=x[j])j=next[j];
        next[++i]=++j;
    }
}
int kmp_count(int x[],int m,int y[],int n)
{
    int i=0,j=0,next[10005];
    kmp_pre(x,m,next);
    while(i<n)
    {
        while(j!=-1&&x[j]!=y[i])j=next[j];
        i++;j++;
        if(j>=m)return i+1-m;///注意這里要減去模式串的長度 } return -1;///如果沒有在主串中找到子串,最后輸出-1表示找不到
}
int a[1000006],b[10004],n,m,t;
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        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]);
        printf("%d\n",kmp_count(b,m,a,n));///直接輸出第一次查找到完整字符串的位置 } } 

注意!

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