HDU4272LianLianKan（dfs）

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

2012 ACM/ICPC Asia Regional Changchun Online

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注意给的顺序是从底部到顶部。

开始想直接模拟，不可行，因为在下面5个里可能有一个或多个相同的元素，每个选择都可能对后面造成影响，因此要dfs。

其中还巧妙的用了map，当然改用set来存然后看里面到底有没有奇数个的元素也是可以的，这点的优化很重要。

#include<stdio.h>
#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<map>

using namespace std;

int vis[1001],a[1001];

int dfs(int n)
{
    int i=0,j=n-1;
    while(n>=0&&vis[n]) --n;
    if(n==-1) return 1;
    if(n==0&&!vis[0]) return 0;
    while(i<=5)
    {
        if(j<0) return 0;
        if(vis[j]) --j;
        if(a[n]==a[j])
        {
            vis[j]=1;
            if(dfs(n-1)) return 1;
            vis[j]=0;
        }
        ++i;
        --j;
    } 
    return 0;
}

int main()
{
    int k,m,q,t,p,n;
    int T;
    map<int,int> mp;
    map<int,int>::iterator it;
    
    while(cin>>n)
    {
        t=0;
        for(int i=0;i<n;++i)
        {
            scanf("%d",&a[i]);
            vis[i]=0;
            mp[a[i]]++;
        }
        
        for(it=mp.begin();it!=mp.end();++it)
        {
            if((*it).second%2)
            {
                t=1;
                break;
            }
        }
        
        if(t)
        {
            cout<<0<<endl;
        }else
        {
            cout<<dfs(n-1)<<endl;
        }

    }
    return 0;
}