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
2012 ACM/ICPC Asia Regional Changchun Online  

Recommend liuyiding     直接暴力DFS。很简单。。。注意细节。还有前面要用map来判断，否则会超时。 听说比较坑的题目。。。好像现在距离取5和6都可以AC了。

#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;
}