Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 53712 Accepted Submission(s): 21540
http://acm.hdu.edu.cn/showproblem.php?pid=1711
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 #include<cstdio> #include<cstring> const int maxn=1e6+4; int next[10000+5]; int a[maxn],b[10000+5]; int t,n,m; int main(){ scanf("%d",&t); while(t--){ scanf("%d%d",&n,&m); int i; for(i=0;i<n;i++){ scanf("%d",&a[i]); } for(i=0;i<m;i++){ scanf("%d",&b[i]); } int j=0,k=-1; next[0]=-1; while(j<m){ if(k==-1||b[k]==b[j]){ ++k; ++j; if(b[j]!=b[k]){ next[j]=k; }else{ next[j]=next[k]; } }else{ k=next[k]; } } i=0,j=0; int flag=0; while(i<n&&j<m){ if(j==-1||a[i]==b[j]){ i++; j++; }else{ j=next[j]; } // printf("%d %d\n",i,j); if(j==m-1&&a[i]==b[j]){ // printf("%d %d\n",i,j); printf("%d\n",i-j+1); flag=1; break; } } if(flag==0){ printf("-1\n"); } } return 0; }
转载于:https://www.cnblogs.com/qqshiacm/p/11506768.html
