hdu 4272 LianLianKan

LianLianKan
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2285 Accepted Submission(s): 701


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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 1 //31MS    5904K    1563 B    C++
 2 /*
 3 
 4     题意：
 5         给出一个栈，与栈顶的的元素距离五以内并且值相等的
 6     元素可以和栈顶元素一起出栈，判断最后栈可否清空 
 7 
 8     状态压缩：
 9         其实这题并不难，应为题目要求离栈顶距离为五的元素，
10     故取10位就可以了，应为每匹配掉一个元素，就会有下一个元素
11     进来，最坏情况1000011111，假如是0开头，状态将会左移一位并加一
12     
13     dp[i][j]表示深度为i状态为j时的情况,0表示空,1表示非空 
14       
15     (不能用贪心解法)
16     9   
17     1 1 2 1 1 1 1 1 2
18 
19 */
20 #include<stdio.h>
21 #include<string.h>
22 #define N 1200
23 int dp[N][N]; //dp[i][j]表示深度为i时状态j的情况 
24 int a[N];
25 int n;
26 int dfs(int depth,int temp)
27 {
28     if(depth<0){
29         if(temp==0) return 1;
30         return 0;
31     }
32     if(dp[depth][temp]!=-1) return dp[depth][temp];
33     int &ret=dp[depth][temp];
34     ret=0;
35     if(!((1<<9)&temp)){ //如果该位已被选去 
36         int ntemp=(temp<<1);
37         if(depth-10>=0) ntemp|=1;
38         ret=dfs(depth-1,ntemp);
39     }else{
40         int cnt=0;
41         int ntemp=(temp^(1<<9));  //去掉栈顶元素,一点定要清零,不然移位会溢出32位 
42         for(int i=1,j=8;i<=9,j>=0;i++,j--){
43             if((1<<j)&ntemp){ //该位未被选过 
44                 cnt++;
45                 if(cnt>5) break;
46                 if(a[depth-i]==a[depth]){
47                     int tmp=(ntemp^(1<<j));
48                     tmp<<=1;
49                     if(depth-10>=0) tmp|=1;
50                     ret=dfs(depth-1,tmp);
51                     if(ret) break;
52                 }
53             }
54         }
55     }
56     return ret;
57 }
58 int main(void)
59 {
60     while(scanf("%d",&n)!=EOF)
61     {
62         for(int i=0;i<n;i++)
63             scanf("%d",&a[i]);
64         if(n&1){
65             puts("0");continue;
66         }
67         memset(dp,-1,sizeof(dp));
68         int temp=0; //初始状态 
69         int j=0;
70         for(int i=n-1;i>=0 && j<=9;i--,j++)
71             temp=(temp<<1)+1;
72         while(j<=9){
73             temp<<=1;j++;
74         }
75         int ans=dfs(n-1,temp);
76         if(ans) puts("1");
77         else puts("0");
78     }
79     return 0;
80 }

 

posted @ 2013-10-17 13:13  heaventouch  阅读(149)  评论(0)  编辑  收藏  举报