HDU 4923 Room and Moor (数学＋单调栈)

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Room and Moor

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 1292 Accepted Submission(s): 418

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

491 1 1 1 1 0 0 1 191 1 0 0 1 1 1 1 140 0 1 140 1 1 1

Sample Output

1.4285711.0000000.0000000.000000

Author

BUPT

Source

2014 Multi-University Training Contest 6

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4923

题目大意：给一个0 1序列，要求构造一个实数序列，(要求bi属于[0, 1]且bi单调递增(非严格递增)) 使得(ai - bi)^2的和最小，输出这个结果

题目分析：首先原来0 1序列的最前面的连续0和最后面的连续1可以忽略，因为序列b中可以构造对应的0 1，使其差为0，接着我们取剩下序列中形如10，1100。。。的子序列，即连续1加上连续0的子序列，尽可能让每一个这样的子序列构造出的结果最小，策略显然是尽量取平均！比如最简单的1100，我们构造的bi对应的值应该是0.5 0.5 0.5 0.5，假设子序列有c0个0，c1个1，平均值为val，则val = (c1 * 1 + c0 * 0) / (c0 + c1)，因此我们可以得到val为c1 / (c0 + c1)，然后维护一个单调栈即可，当当前的val比栈顶的val小时，取出栈顶元素，将这段序列与栈顶序列中和，再入栈，中和的方式就是重新计算val值，最后将栈中的元素纷纷弹出并累计c1 * (1 - val)^2 + c0 * val^2
比如样例1：1 1 1 1 1 0 0，val = 5 / 7，则ans = 5 * (1 - 5/7)^2 + 2 * (5/7)^2

#include <cstdio>#include <cstring>#include <stack>using namespace std;int const MAX = 1e5 + 5;struct DIVIDE{    int c0, c1;    double val;}d[MAX];int a[MAX];stack <DIVIDE> stk;int main(){    int T, n;    scanf("%d", &T);    while(T--)    {        double ans = 0;        memset(a, 0, sizeof(a));        scanf("%d", &n);        int st = 1, ed = n;        for(int i = 1; i <= n; i++)            scanf("%d", &a[i]);        while(ed >= 1 && a[ed] == 1)            ed --;        while(st <= n && a[st] == 0)            st ++;        int cnt = 0;        for(int i = st; i <= ed; i++)        {            if(a[i] == 1)            {                if(a[i - 1] == 0)                {                    cnt ++;                    d[cnt].c0 = d[cnt].c1 = 0;                }                d[cnt].c1 ++;            }            else                d[cnt].c0 ++;        }        for(int i = 1; i <= cnt; i++)            d[i].val = (d[i].c1 * 1.0) / ((d[i].c1 + d[i].c0) * 1.0);        for(int i = 1; i <= cnt; i++)        {            if(stk.empty() || d[i].val >= stk.top().val)                stk.push(d[i]);            else            {                while(!stk.empty() && d[i].val < stk.top().val)                {                    d[i].c0 += stk.top().c0;                    d[i].c1 += stk.top().c1;                    d[i].val = (d[i].c1 * 1.0) / ((d[i].c1 + d[i].c0) * 1.0);                    stk.pop();                }                i --;            }        }        while(!stk.empty())        {            int c1 = stk.top().c1;            int c0 = stk.top().c0;            double val = stk.top().val;            ans += c1 * (1.0 - val) * (1.0 - val) + c0 * val * val;            stk.pop();        }        printf("%.6f\n", ans);    }}

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