BUPT OJ183 Longest Subsequence

题目描述

 Recently,Nero is researching a kind of sequence problem related to the sphere.This kind of problem ressembles the Longest Ascending Subsequence.
For instance,Given a sequence,can you find the longest subsequence in it?
Following constraints should be meeted:
1:the elements in the subsequence should be totally equal.
2:the interval between the adjacent elements in the subsequence shouldn't be greater than one(in the hint,we'll show clearly what the interval is)

输入格式

 T indicating the number of test cases
 Each case begins with a interger n(0<=n<=1000),representing the length of sequence
 Following are n integers,Pi stands for the value of each element.

输出格式

 One line,contains a integer indicating the length of the longest subsequence.

输入样例

  2  6  1 2 2 1 1 1  8  1 2 2 2 1 1 1 1

输出样例

34

题干提到的LAS真够坑的... 明明没有关系的, sample给的偏偏就是模糊不清, 理解为LAS也可以得出相同结果

其实简单的很, 没理解题目意思, 结果自然是WA了以后立刻就发现了, 理解了就好简单...

事实上题目意思是要找一个最长的相同元素构成的序列, 使得相邻两项的位置差不超过1...也就是最多可以跳一位取下一项

思路是简单的dp, 毫无技术含量......能不跳就不跳, 其实说是贪心也对? 撒~~

/*USER_ID: test#birdstormPROBLEM: 183SUBMISSION_TIME: 2014-03-08 01:07:10*/#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#define MAX(x,y) (x)>(y)?(x):(y)#define For(i,m,n) for(i=m;i<n;i++)#define MAXN 1005 int a[MAXN], dp[MAXN]; main(){    int i, j, n, t, len, max;    scanf("%d",&t);    while(t--){        len=1; max=0;        dp[0]=1;        scanf("%d%d",&n,&a[0]);        For(i,1,n){            scanf("%d",&a[i]);            if(a[i]==a[i-1]) dp[i]=dp[i-1]+1;            else if(a[i]==a[i-2]) dp[i]=dp[i-2]+1;            else dp[i]=1;        }        For(i,0,n) max=MAX(max,dp[i]);        printf("%d\n",max);    }    return 0;}