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:

 /*************************************************************************
     > File Name: 4923.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年08月08日 星期五 22时27分38秒
     > Propose:
  ************************************************************************/

 #include <cmath>
 #include <string>
 #include <cstdio>
 #include <fstream>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const double eps = 1e-;
 const int maxn = ;
 int n;
 int a[maxn], st[maxn];

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

 // unite i to j
 void unite(int i, int j) {
       A[j].l = A[i].l;
     A[j].x = A[j].one + A[i].one + 0.0;
     A[j].x /= A[j].r - A[j].l + ;
     A[j].one = A[i].one + A[j].one;
 }

 double solve() {
       int len = ;
     for (int i = ; i <= n; i++) {
           A[len].x = a[i] + 0.0;
         A[len].l = i;
         i++;
         while (i <= n && a[i] == a[i-]) i++;
         A[len].r = i - ;
         i--;
         if (a[i]) A[len].one = A[len].r - A[len].l + ;
         else A[len].one = ;
         len++;
     }
     int top = ;
     for (int i = ; i < len; i++) {
           while (top && A[i].x < A[st[top-]].x) {
               unite(st[top-], i);
             top--;
         }
         st[top++] = i;
     }
     double ans = 0.0;
     for (int i = ; i < top; i++) {
           for (int j = A[st[i]].l; j <= A[st[i]].r; j++) {
               ans += (a[j] - A[st[i]].x) * (a[j] - A[st[i]].x);
         }
     }
     return ans + eps;
 }

 int main(void) {
       int t;
     scanf("%d", &t);
     while (t--) {
           scanf("%d", &n);
         for (int i = ; i <= n; i++) scanf("%d", a + i);
         printf("%.6f\n", solve());
     }
     return ;
 }