HDU 1711 Number Sequence(KMP)

Mr_Treeeee發表於2020-04-06
KMP

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27986    Accepted Submission(s): 11777


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





 



Source



HDU 2007-Spring Programming Contest 



 


KMP，返回第一次出現的地址。只是不是字串而是int陣列而已。


#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
int x[1000000+5];
int y[10000+5];
int l1,l2;
int nxt[10000+5];
void kmpnext()
{
    int i=0,j=-1;
    nxt[0]=-1;
    while(i<l2)
    {
        while(j!=-1&&y[i]!=y[j])
            j=nxt[j];
        nxt[++i]=++j;
    }
    
}
int kmp()
{
    kmpnext();
    int i=0,j=0;
    while(i<l1)
    {
        while(j!=-1&&x[i]!=y[j])
            j=nxt[j];
        i++;j++;
        if(j>=l2)
        {
            return i-l2+1;
        }
    }
    return -1;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(nxt,0,sizeof nxt);
        scanf("%d %d",&l1,&l2);
        for(int i=0;i<l1;i++)
        {
            scanf("%d",&x[i]);
        }
        for(int i=0;i<l2;i++)
        {
            scanf("%d",&y[i]);
        }
        printf("%d\n",kmp());
        
    }

}









                                    




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