【HDU】4923 Room and Moor(2014多校第六场1003)

Room and Moor

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 263    Accepted Submission(s): 73

Problem Description

PM Room defines a sequence A = {A₁, A₂,..., A_N}, each of which is either 0 or 1. In order to beat him, programmer Moor has to construct another sequence B = {B₁, B₂,... , B_N} of the same length, which satisfies that:

 

Input

The input consists of multiple test cases. The number of test cases T(T<=100) occurs in the first line of input.

For each test case:
The first line contains a single integer N (1<=N<=100000), which denotes the length of A and B.
The second line consists of N integers, where the ith denotes A_i.

 

Output

Output the minimal f (A, B) when B is optimal and round it to 6 decimals.

 

Sample Input

4

9

1 1 1 1 1 0 0 1 1

9

1 1 0 0 1 1 1 1 1

4

0 0 1 1

4

0 1 1 1

 

Sample Output

1.428571

1.000000

0.000000

0.000000

 

Source

2014 Multi-University Training Contest 6

 

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题意：很容易就读懂了。

题解：首先去掉前导零和最后的1，相当于把整个序列分成几个区间，每部分以1开头，0结尾，即如1 0   1 1 0 0等，可知对于每一个区间，要取得最小值，那这个部分所有的值即对应的这个区间内的平均数，如果这个平均数和前面一个区间的相比较大，就压入栈，否则将栈里的元素顶出，并与当前区间合并求平均数……知道比前面的大为止，最后求出每个区间的对应的Seg(ai - bi)^2 就可以了。至于为什么。。。。说实话全是YY，居然A掉了。。

 

AC代码如下：

 

 #include <cstdio>
 #include <cstring>
 #include <stack>
 using namespace std;

 #define eps 0.00000001
 const int LEN = ;

 int arr[LEN];
 struct line
 {
     int l, r, sum;
     double rate;
 };

 stack<line> s;

 int main()
 {
     int T, n;
     line tmp;
     scanf("%d", &T);
     while(T--){
         scanf("%d", &n);
         for(int i = ; i < n; i++)
             scanf("%d", arr+i);
         int h = ;
         while(arr[h] == )
             h++;
         int k = n - ;
         while(arr[k] == )
             k--;
         for(int i = h; i <= k; i++){
             if (i == h || i > h && arr[i-] ==  && arr[i] == ){
                 tmp.l = i;
                 tmp.sum = ;
             }
             if (i < k && arr[i] ==  && arr[i+] ==  || i == k){
                 tmp.r = i;
                 //printf("l = %d, r = %d\n", tmp.l, tmp.r);
                 tmp.rate = tmp.sum * 1.0 / ((tmp.r - tmp.l + ) * 1.0);
                 //printf("rate=%f\n", tmp.rate);
                 while(true){
                     if (s.empty() || s.top().rate - tmp.rate < eps){
                         s.push(tmp);
                         break;
                     }
                     if (s.top().rate - tmp.rate > eps){
                         tmp.l = s.top().l;
                         tmp.sum += s.top().sum;
                         tmp.rate = tmp.sum*1.0 / ((tmp.r - tmp.l + )*1.0);
                         s.pop();
                     }
                 }
             }
             if (arr[i] == )
                 tmp.sum++;
         }
         double ans = ;
         while(!s.empty()){
             ans += (( - s.top().rate) * ( - s.top().rate) * s.top().sum + s.top().rate * s.top().rate * (s.top().r - s.top().l +  - s.top().sum));
             s.pop();
         }
         printf("%f\n", ans);
     }
     return ;
 }