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hdu4923

2024-07-16 10:07:13   222人阅读

Room and Moor

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 557    Accepted Submission(s): 159


Problem Description
PM Room defines a sequence A = {A1, A2,..., AN}, each of which is either 0 or 1. In order to beat him, programmer Moor has to construct another sequence B = {B1, B2,... , BN} 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 Ai.
 

 

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
 
思路：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。
 1 #include <cstdio> 2 #include <cstring> 3 #include <stack> 4 #define eps 0.00000001 5 using namespace std; 6  7 const int LEN = 100010; 8 int arr[LEN]; 9 struct line10 {11     int l, r, sum;12     double rate;13 };14 stack<line> s;15 int main()16 {17     int T, n;18     line tmp;19     scanf("%d", &T);20     while(T--)21     {22         scanf("%d", &n);23         for(int i = 0; i < n; i++)24             scanf("%d", &arr[i]);25         int h = 0;26         while(arr[h] == 0)27             h++;28         int k = n - 1;29         while(arr[k] == 1)30             k--;31         for(int i = h; i <= k; i++)32         {33             if (i == h || i > h && arr[i-1] == 0 && arr[i] == 1)34             {35                 tmp.l = i;36                 tmp.sum = 0;37             }38             if (i < k && arr[i] == 0 && arr[i+1] == 1 || i == k)39             {40                 tmp.r = i;41                 tmp.rate = tmp.sum * 1.0 / ((tmp.r - tmp.l + 1) * 1.0);42                 while(true)43                 {44                     if (s.empty() || s.top().rate - tmp.rate < eps)45                     {46                         s.push(tmp);47                         break;48                     }49                     if (s.top().rate - tmp.rate > eps)50                     {51                         tmp.l = s.top().l;52                         tmp.sum += s.top().sum;53                         tmp.rate = tmp.sum*1.0 / ((tmp.r - tmp.l + 1)*1.0);54                         s.pop();55                     }56                 }57             }58             if (arr[i] == 1)59                 tmp.sum++;60         }61         double ans = 0;62         while(!s.empty())63         {64             ans += ((1 - s.top().rate) * (1 - s.top().rate) * s.top().sum + s.top().rate * s.top().rate * (s.top().r - s.top().l + 1 - s.top().sum));65             s.pop();66         }67         printf("%f\n", ans);68     }69     return 0;70 }
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