Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30893 Accepted Submission(s): 12981
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
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lcy
题解：
①KMP入门吧。
②贴一个板子，以前总是不熟练(打自己一下,啊!)。
#include<stdio.h>
#define go(i,a,b) for(int i=a;i<=b;i++)
int n,m,f[10003],j,C,T[1000004],P[10003];
int main()
{
scanf("%d",&C);
while(C--&&scanf("%d%d",&n,&m))
{
go(i,0,n-1)scanf("%d",T+i);
go(i,0,m-1)scanf("%d",P+i);f[0]=f[1]=0;
go(i,1,m-1){j=f[i];while(j&&P[i]!=P[j])j=f[j];f[i+1]=P[i]==P[j]?j+1:0;}j=0;
go(i,0,n-1){while(j&&P[j]!=T[i])j=f[j];j+=P[j]==T[i];
if(j==m){printf("%d
",i-j+2);goto _;}}puts("-1");_:;
}
return 0;
}//Paul_Guderian
Birds don't just fly they fall down and get up,
Nobody learns without getting it won.——————<Zootopia>