Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6296 Accepted Submission(s): 2845
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
Recommend lcy
题意：给两个数字序列，判断第一个是否包含了第二个数列。
分析：KMP算法的典型运用。
KMP算法（详细资料参考：http://www.cnblogs.com/dolphin0520/archive/2011/08/24/2151846.html）
#include<iostream> using namespace std; #define MAXN 1000000 #define MAXM 10000 int a[MAXN],b[MAXM]; int next[MAXN]; int n,m; void getNext(int p[],int next[]) { int j,k; next[0]=-1; j=0; k=-1; while(j<m-1) { if(k==-1||p[j]==p[k])//匹配的情况下,p[j]==p[k] { j++; k++; next[j]=k; } else//p[j]!=p[k] k=next[k]; } } int KMPMatch(int s[],int p[]) { int i,j; i=0; j=0; getNext(p,next); while(i<n) { if(j==-1||s[i]==p[j]) { i++; j++; } else { j=next[j];//消除了指针i的回溯 } if(j==m) return i-m+1; } return -1; } int main() { int t,i,j; scanf("%d",&t); while(t--) { scanf("%d %d",&n,&m); for(i=0;i<n;i++) scanf("%d",&a[i]); for(i=0;i<m;i++) scanf("%d",&b[i]); printf("%d\n",KMPMatch(a,b)); } return 0; }
转载于:https://www.cnblogs.com/Su-Blog/archive/2012/09/13/2684021.html
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