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HDU-4272 LianLianKan
http://acm.hdu.edu.cn/showproblem.php?pid=4272
据说是状态压缩,+dfs什么什么的,可我这样也过了,什么算法都是浮云 ,暴力才是王道。我也归类为状态压缩,可以用状态压缩来做。
LianLianKan
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2482 Accepted Submission(s): 773
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.
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)
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
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 int main() 6 { 7 int n,j,i,k,a[1005],v[1005]; 8 while(~scanf("%d",&n)) 9 {10 for(i=1;i<=n;i++)11 scanf("%d",&a[i]);12 memset(v,0,sizeof(v));13 int cnt=0;14 int flag=0;15 while(cnt<n)16 {17 for(i=1;i<=n;i++)18 {19 if(v[i]==1)20 continue;21 k=0;22 for(j=i+1;j<=n&&k<=5;j++)23 {24 if(v[j]==1)25 continue;26 if(a[i]==a[j])27 {28 v[i]=v[j]=1;29 break;30 }31 k++;32 }33 }34 cnt++;35 }36 for(i=1;i<=n;i++)37 {38 if(v[i]==0)39 {40 printf("0\n");41 break;42 }43 }44 if(i>n)45 printf("1\n");46 47 }48 return 0;49 }
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