HDU 1711 Number Sequence（KMP）

最新推荐文章于 2019-07-14 13:24:47 发布

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最新推荐文章于 2019-07-14 13:24:47 发布

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Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22717 Accepted Submission(s): 9717

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

KMP。。。

代码如下：

#include<cstdio>
#include<cstring>
#define p 2000000
int next[10005]={0},zhu[1000005]={0},mo[10005]={0};
void getnext()
{
    int i=0,j=-1;
    next[0]=-1;
    while(mo[i]!=p)
    {
        while(j!=-1&&mo[i]!=mo[j]) j=next[j];
        if(mo[++i]!=mo[++j]) next[i]=j;
        else next[i]=next[j];
    }
}
int KMP(int len)
{
    int i=0,j=0;
    next[0]=-1;
    while(zhu[i]!=p)
    {
        if(j!=-1&&mo[j]==p) break;
        while(j!=-1&&zhu[i]!=mo[j]) j=next[j];
        ++i,++j;
    }
    return mo[j]==p?i-j+1:-1;
}
int main()
{
    int i,j,m,n,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&m);
        for(i=0;i<n;++i) scanf("%d",&zhu[i]);
        for(i=0;i<m;++i) scanf("%d",&mo[i]);
        zhu[n]=p;
        mo[m]=p;
        getnext();
        printf("%d\n",KMP(m));
    }
    return 0;
}