【HDU】4272 LianLianKan（DFS）

LianLianKan

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

直接暴力DFS，超简单，无话可说。

#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;}

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大多数人想要改造这个世界，但却罕有人想改造自己。

世上没有绝望的处境，只有对处境绝望的人。

                                              ————By slience

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