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-12;
const int maxn = 100002;
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 + 1;
    A[j].one = A[i].one + A[j].one;
}

double solve() {
      int len = 1;
    for (int i = 1; i <= n; i++) {
          A[len].x = a[i] + 0.0;
        A[len].l = i;
        i++;
        while (i <= n && a[i] == a[i-1]) i++;
        A[len].r = i - 1;
        i--;
        if (a[i]) A[len].one = A[len].r - A[len].l + 1;
        else A[len].one = 0;
        len++;
    } 
    int top = 0;
    for (int i = 1; i < len; i++) {
          while (top && A[i].x < A[st[top-1]].x) {
              unite(st[top-1], i);
            top--;
        }
        st[top++] = i;
    }
    double ans = 0.0;
    for (int i = 0; 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 = 1; i <= n; i++) scanf("%d", a + i);
        printf("%.6f
", solve());
    }
    return 0;
}