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[],l1,l2,sl;
 int S[],T[];
 int solve()
 {
     int i,j;
     nex[]=nex[]=;
     for(i=;i<l2;++i)
     {
         j=nex[i];
         while(j&&T[i]!=T[j])j=nex[j];
         nex[i+]=T[i]==T[j]?j+:;
     }
     j=;
     for(i=;i<l1;++i)
     {
         while(j&&S[i]!=T[j])j=nex[j];
         if(S[i]==T[j]){
             j++;
             if(j==l2)return i-l2+;
         }
     }
     return -;
 }
 int main()
 {
     int i,j,t,p,x,tmp;
     cin>>t;
     while(t--){p=;M.clear();
         scanf("%d%d",&l1,&l2);
         for(i=;i<l1;++i)
         {
             scanf("%d",&x);
             tmp=M[x];
             if(!tmp){M[x]=S[i]=++p;}
             else S[i]=tmp;
         }
         for(i=;i<l2;++i)
         {
             scanf("%d",&x);
             tmp=M[x];
             if(!tmp){M[x]=T[i]=++p;}
             else T[i]=tmp;
         }
         cout<<solve()<<endl;;
     }
     return ;
 }