Hdu 4923（单调栈）

题目链接

Room and Moor

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 842    Accepted Submission(s): 250

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

 Accepted Code:

 1 /*************************************************************************  2  > File Name: 4923.cpp  3  > Author: Stomach_ache  4  > Mail: [email protected]  5  > Created Time: 2014年08月08日 星期五 22时27分38秒  6  > Propose:  7  ************************************************************************/

 8 

 9 #include <cmath>

10 #include <string>

11 #include <cstdio>

12 #include <fstream>

13 #include <cstring>

14 #include <iostream>

15 #include <algorithm>

16 using namespace std; 17 

18 const double eps = 1e-12; 19 const int maxn = 100002; 20 int n; 21 int a[maxn], st[maxn]; 22 

23 struct node { 24     double x; //区间均值 25     int l, r, one; //区间范围及1的个数 26 }A[maxn]; 27 

28 // unite i to j

29 void unite(int i, int j) { 30       A[j].l = A[i].l; 31     A[j].x = A[j].one + A[i].one + 0.0; 32     A[j].x /= A[j].r - A[j].l + 1; 33     A[j].one = A[i].one + A[j].one; 34 } 35 

36 double solve() { 37       int len = 1; 38     for (int i = 1; i <= n; i++) { 39           A[len].x = a[i] + 0.0; 40         A[len].l = i; 41         i++; 42         while (i <= n && a[i] == a[i-1]) i++; 43         A[len].r = i - 1; 44         i--; 45         if (a[i]) A[len].one = A[len].r - A[len].l + 1; 46         else A[len].one = 0; 47         len++; 48  } 49     int top = 0; 50     for (int i = 1; i < len; i++) { 51           while (top && A[i].x < A[st[top-1]].x) { 52               unite(st[top-1], i); 53             top--; 54  } 55         st[top++] = i; 56  } 57     double ans = 0.0; 58     for (int i = 0; i < top; i++) { 59           for (int j = A[st[i]].l; j <= A[st[i]].r; j++) { 60               ans += (a[j] - A[st[i]].x) * (a[j] - A[st[i]].x); 61  } 62  } 63     return ans + eps; 64 } 65 

66 int main(void) { 67       int t; 68     scanf("%d", &t); 69     while (t--) { 70           scanf("%d", &n); 71         for (int i = 1; i <= n; i++) scanf("%d", a + i); 72         printf("%.6f\n", solve()); 73  } 74     return 0; 75 }