HDU 1711 kmp+离散化

http://acm.hdu.edu.cn/showproblem.php?pid=1711

    

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30591    Accepted Submission(s): 12870

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

     把数字离散化后当作字符来处理就是kmp了，注意模板是从下标0开始的我一直输入时从1开始导至答案不对。

   

#include<cstring>
#include<cstdio>
#include<map>
#include<iostream>
using namespace std;
map<int,int>M;
int nex[10005],l1,l2,sl;
int S[1000005],T[10005];
int solve()
{
    int i,j;
    nex[0]=nex[1]=0;
    for(i=1;i<l2;++i)
    {
        j=nex[i];
        while(j&&T[i]!=T[j])j=nex[j];
        nex[i+1]=T[i]==T[j]?j+1:0;
    }
    j=0;
    for(i=0;i<l1;++i)
    {
        while(j&&S[i]!=T[j])j=nex[j];
        if(S[i]==T[j]){
            j++;
            if(j==l2)return i-l2+2;
        }
    }
    return -1;
}
int main()
{
    int i,j,t,p,x,tmp;
    cin>>t;
    while(t--){p=0;M.clear();
        scanf("%d%d",&l1,&l2);
        for(i=0;i<l1;++i)
        {
            scanf("%d",&x);
            tmp=M[x];
            if(!tmp){M[x]=S[i]=++p;}
            else S[i]=tmp;
        }
        for(i=0;i<l2;++i)
        {
            scanf("%d",&x);
            tmp=M[x];
            if(!tmp){M[x]=T[i]=++p;}
            else T[i]=tmp;
        }
        cout<<solve()<<endl;;
    }
    return 0;
}