# LianLianKan

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 840    Accepted Submission(s): 280

Problem Description
I like playing game with my friend, although sometimes looks pretty naive. Today I invent a new game called LianLianKan. The game is about playing on a number stack.
Now we have a number stack, and we should link and pop the same element pairs from top to bottom. Each time, you can just link the top element with one same-value element. After pop them from stack, all left elements will fall down. Although the game seems to be interesting, it's really naive indeed. To prove I am a wisdom among my friend, I add an additional rule to the game: for each top element, it can just link with the same-value element whose distance is less than 6 with it.
Before the game, I want to check whether I have a solution to pop all elements in the stack.

Input
There are multiple test cases.
The first line is an integer N indicating the number of elements in the stack initially. (1 <= N <= 1000)
The next line contains N integer ai indicating the elements from bottom to top. (0 <= ai <= 2,000,000,000)

Output
For each test case, output “1” if I can pop all elements; otherwise output “0”.

Sample Input
2 1 1 3 1 1 1 2 1000000 1

Sample Output
1 0 0

Source

Recommend
liuyiding

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<map>
#include<algorithm>
using namespace std;
const int MAXN=1010;
int a[MAXN];
bool used[MAXN];
int dfs(int n)
{
while(n>0&&used[n])n--;
if(n==0)return 1;
if(n==1)return 0;
int i=0;
int j=n-1;
for(;i<=5;)//这里取i<5和i<=5都可以AC
{
if(j<=0)return 0;//没有找到相等的
if(used[j])
{
j--;
continue;
}
if(a[n]==a[j])
{
used[j]=true;
if(dfs(n-1)) return 1;
used[j]=false;
}
i++;
j--;
}
return 0;
}
map<int,int>mp;
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
mp.clear();
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
used[i]=false;
mp[a[i]]++;
}
if(n&1)
{
printf("0\n");
continue;
}
int t=1;
//加个map判断就是0ms,否则就是TLE
map<int,int>::iterator it;
for(it=mp.begin();it!=mp.end();it++)
{
if((it->second)%2==1)
{
t=0;
break;
}
}
if(t==0)
{
printf("0\n");
continue;
}
printf("%d\n",dfs(n));
}
return 0;
}

posted on 2012-09-11 09:45 kuangbin 阅读(...) 评论(...) 编辑 收藏

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