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 }

Hdu 4923（单调栈）