zoukankan      html  css  js  c++  java
  • hdoj 1711 Number Sequence【求字串在母串中第一次出现的位置】

    Number Sequence

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 15017    Accepted Submission(s): 6585


    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<stdio.h>
    #include<string.h>
    #define MAX 10010
    int n,m;
    int b[MAX],a[MAX*100];
    int next[MAX];
    void getmap()
    {
    	int i;
    	for(i=0;i<n;i++)
    	scanf("%d",&a[i]);
    	for(i=0;i<m;i++)
    	scanf("%d",&b[i]);
    }
    void getfail()
    {
    	int i,j;
    	next[0]=next[1]=0;
        for(i=1;i<m;i++)
        {
        	j=next[i];
        	while(j&&b[i]!=b[j])
        	    j=next[j];
        	next[i+1]=b[i]==b[j]?j+1:0;
    	}
    }
    void kmp()
    {
    	int i,j=0;
    	int ok=0;
    	for(i=0;i<n;i++)
    	{
    		while(j&&a[i]!=b[j])
    		    j=next[j];
    		if(a[i]==b[j])
    		    j++;
    		if(j==m)
    		{
    			ok=1;
    			printf("%d
    ",i-m+2);//难点在这里 
    			return ;
    		}
    	}
    	if(!ok)
    	printf("-1
    "); 
    }
    int main()
    {
    	int i,t;
    	scanf("%d",&t);
    	while(t--)
    	{
    		scanf("%d%d",&n,&m);
    		getmap();
    		getfail();
    		kmp();
    	}
    	return 0;
    }
    

      

  • 相关阅读:
    【转】SVN与Git比较
    我遇到了Hibernate异常
    使用 Eclipse 远程调试 Java 应用程序
    linux显示桌面快捷键设置
    Ubuntu共享WiFi(AP)给Android方法
    用zd1211+Ubuntu 10.04实现的AP
    Ubuntu开机自动禁用无线网络
    戴尔大力宣传Ubuntu 对比与Windows的差异
    【SSH进阶之路】Spring的AOP逐层深入——采用注解完成AOP(七)
    【SSH进阶之路】Spring的AOP逐层深入——AOP的基本原理(六)
  • 原文地址:https://www.cnblogs.com/tonghao/p/4711898.html
Copyright © 2011-2022 走看看