HDU 4923 Room and Moor （多校第六场C题） 单调栈

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

能够分析出 所求的区间 也就是 从第一个为1開始的到最后一个0结束。每段都是形如111...111000...000这样先为1后为0 的小区间里 B值都是水平的。那么先求出当前一零区间的最优值，假设当前的高度最优值大于单调增栈里 栈首元素的高度，那么能够直接入栈，假设小于，就取出栈首元素与当前区间进行合并，再次入栈。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stack>
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
using namespace std;
typedef long long LL;
const int maxn = 100005;
const int mod = 1000000007;
int t, n, a[maxn];
struct C {
    int num1, num0;
    double res, h;
};
double Cu(double x, double y)  {
    return x*y/(x+y);
}
double Ch(double x, double y) {
    return x/(x+y);
}
int main()
{
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        int st = 1, en = n, i, j, k;
        for(i = 1; i <= n; i++) scanf("%d", &a[i]);
        for(i = 1; i <= n && a[i] == 0; i++) ; st = i;
        for(i = n; i >= 1 && a[i] == 1; i--) ; en = i;
        stack <C> zhan;
        for(i = st; i <= en; ) {
            int u = 0, d = 0;
            for(j = i; j <= en; j++) {
                if(a[j] == 0) break;
                u++;
            }
            for(k = j; k <= en; k++) {
                if(a[k] == 1) break;
                d++;
            }
            C aa;
            aa.num0 = d;
            aa.num1 = u;
            aa.res = Cu(u, d);
            aa.h = Ch(u, d);
            while(!zhan.empty() && zhan.top().h > aa.h) {
                C tmp = zhan.top();
                zhan.pop();
                aa.num1 = tmp.num1 + aa.num1;
                aa.num0 = tmp.num0 + aa.num0;
                aa.res = Cu(aa.num1, aa.num0);
                aa.h = Ch(aa.num1, aa.num0);
            }
            zhan.push(aa);
            i = k;
        }
        double sum = 0;
        while(!zhan.empty()) {
            C tmp = zhan.top();
            zhan.pop();
            sum += tmp.res;
        }
        printf("%.6lf\n", sum);
    }
    return 0;
}