HDU 4272 LianLianKan

LianLianKan

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4093    Accepted Submission(s): 1231

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

21 131 1 121000000 1

 

Sample Output

100

 

题目大意：连连看，不超过5的距离的两个相同的元素可以消除

模拟一下

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int num[1010];int book[1010];int main(){//freopen("in.txt","r",stdin);int n;while(cin>>n){memset(book,0,sizeof(book));for(int i=1;i<=n;i++) scanf("%d",&num[i]);int cnt=0;while(cnt<n){int i;for(i=1;i<=n;i++){if(book[i]) continue;int step=0;for(int j=i+1;j<=n&&step<=5;j++){if(book[j]) continue;else{if(num[j]==num[i]){book[j]=book[i]=1;break;}}step++;}}//第一次消除时可能会因为两个元素间有很多可以消除但是//并未消除的元素，消除中间元素以后就可以消除了，但是因为每次//循环都是往后边的元素找，所以我们要多次重新开始循环//确保可以消除的都消除//比如3 5 5 6 6 7 7 8 8 3 cnt++;}int i;for(i=1;i<=n;i++){if(!book[i]){cout<<0<<endl;break;}}if(i>n) cout<<1<<endl;}}