Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3681 Accepted Submission(s): 1101
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
http://acm.hdu.edu.cn/showproblem.php?pid=4272
题意：
题意：长度为n（n<=1000）的栈，栈顶元素可以与下面1~5个数中相同的元素消去，问最后能都完全消去。
题解：
状态压缩 dp 如何判断一个 物品 是否可以 被删除 ，首先 最坏的 情况是 2 0 0 0 0 1 1 1 1 2 假如我们要消除 栈顶的 2 ，0表示已经被删除了。
辅助解释：假设2前面有一个3，那么3能够消除的最远位置为最后一个0处；
我们要 知道 包括 本身在内的 10 个 数位
定义dp[h][i]表示高度为h，状态为i时能否全部消除； 二进制位1表示未消除，0表示当前位已消除；
/**定义dp[h][i]表示高度为h，状态为i时能否全部消除； 二进制位1表示未消除，0表示当前位已消除；*/#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<set>#include<algorithm>#include<vector>#include<map>#include<queue>#include<cmath>#include<cctype>#include<stack>#include<iomanip>using namespace std;const int maxn = 1200;int dp[maxn][maxn];int a[maxn];int dfs(int h,int c){ if(h < 0){
if(c) return 0;//
else return 1;// 结果为0，表明可以完全消除；
}
if(dp[h][c]!=-1) return dp[h][c];
int &res = dp[h][c];
res = 0;
int t = c&(1<<9);//获得定点的位，以判断是否已经被消除；
if(t){//未被消除；
c = c^(1<<9);//把顶去掉；
int cnt = 0;
for(int i = 1,j = 8;i <= 9&&j>=0; i++,j--){
if(c&(1<<j)){//如果当前位置未去掉；
cnt++;
if(cnt>5) break;//控制距离小于6；
if(a[h] == a[h-i]){//如果和顶相等；去掉当前位；
int t = c^(1<<j);
t <<= 1;
if(h-10>=0) t += 1;
res = dfs(h-1,t);
// else res = dfs(h-1,t);
if(res) break;
}
}
}
}else
{//已经被消除时，脱栈顶；
c <<= 1;
if(h-10>=0) res = dfs(h-1,c+1);//h>=10表明可以继续下移；
else res = dfs(h-1,c);
}
return res;
}int main(){ int n;
while(scanf("%d",&n)==1)
{
for(int i = 0; i < n; i++){
scanf("%d",&a[i]);
}
if(n%2){//数量为奇数个，肯定不可以一一配对完全消除；
printf("0\n"); continue;
}
memset(dp,-1,sizeof dp);
int c = 0, i, j;
for(i = n-1, j = 0; i >= 0&&j<=9; i--,j++){//初始化二进制111111111；
c = (c<<1)+1;
}
while(j<=9){//n<9时，补0；只是为了方便同一处理；
c <<= 1;
j++;
}
printf("%d\n",dfs(n-1,c));
}
return 0;
}