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HDU_1711_初识KMP算法
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22568 Accepted Submission(s): 9639
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
KMP算法还需多理解。
#include<iostream>#include<cstdio>#include<cstring>#include<stdlib.h>#include<algorithm>#include<cmath>using namespace std;int n,m;int N[1000005],M[10005],Pi[10005];void preFix(){ memset(Pi,0,sizeof(Pi)); int k=0; for(int q=2;q<=m;q++) { while(k>0&&M[k+1]!=M[q]) k=Pi[k]; if(M[k+1]==M[q]) k++; Pi[q]=k; }}int KMP(){ preFix(); int q=0; for(int i=1;i<=n;i++) { while(q>0&&M[q+1]!=N[i]) q=Pi[q]; if(M[q+1]==N[i]) q++; if(q==m) return i-m+1; } return -1;}
int main(){ int t; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) scanf("%d",&N[i]); for(int j=1;j<=m;j++) scanf("%d",&M[j]); int ans=KMP(); printf("%d\n",ans); } return 0;}
HDU_1711_初识KMP算法
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