【杭电oj1711】Number Sequence
来源:互联网 发布:魅影剃刀视频数据 编辑:程序博客网 时间:2024/06/06 05:30
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22047 Accepted Submission(s): 9420
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
213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1
Sample Output
6-1
Source
HDU 2007-Spring Programming Contest
Recommend
lcy | We have carefully selected several similar problems for you: 1358 3336 1686 3746 1251
经过此题(虽是模板题)发现我这菜鸟还是对KMP算法没有深入的理解,掌握不好,处处出错。看来要继续深层次理解KMP算法啊。。。。。
#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<algorithm>using namespace std;const int N = 1e6+10;const int M = 1e4+5;int p[N],s[N];int ans,n,m;void getNext(int *s,int *next) {int j,k;next[0]=-1;j=0,k=-1;while(j<m-1) {if(k==-1||s[j]==s[k]) {j++;k++;//if(s[j]!=s[k])next[j]=k;//else//next[j]=next[j];} elsek=next[k];}}int KMPMatch(int *p,int *s) {int i,j;int next[M];i=0,j=0;getNext(s,next); //忘加了//printf("--\n");while(i<n) {if(j==-1||p[i]==s[j]) {j++;i++;//printf("---\n");} else {j=next[j];}if(j==m)return i-m+1;}return -1;}int main() {int T;scanf("%d",&T);while(T--) {scanf("%d%d",&n,&m);for(int i=0; i<n; i++) {scanf("%d",&p[i]);}for(int j=0; j<m; j++) {scanf("%d",&s[j]);}ans=KMPMatch(p,s);printf("%d\n",ans);}return 0;}
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1711
0 0
- 【杭电oj1711】Number Sequence
- [杭电]Number Sequence
- 杭电acm1005 Number Sequence
- 杭电1005 Number Sequence
- 杭电1711 Number Sequence
- 杭电 1711 Number Sequence
- 杭电 1005 Number Sequence
- 杭电 1005 Number Sequence
- 杭电1005 Number Sequence
- 杭电1005Number Sequence
- 杭电1711Number Sequence
- 杭电1005-Number Sequence
- 杭电hdu1005 Number Sequence
- 【杭电1005】Number Sequence
- 【杭电】 1005 Number Sequence
- 杭电ACM1005-Number Sequence
- 杭电1005 Number Sequence
- 杭电problem1005-Number Sequence
- GetFileContents
- ssh和scp(未完)
- Https协议连接过程
- CodeForces-607B Zuma
- Eigen 3.3beta2
- 【杭电oj1711】Number Sequence
- 类和接口的继承与实现
- C++中的explicit关键字
- 数据结构实验图论一:基于邻接矩阵的广度优先搜索遍历
- 图的深度遍历
- qt采用opengl显示yuv视频数据
- MySQL中索引的操作
- 二叉排序树
- PAT甲级练习题A1022. Digital Library (30)