hdu 4923 Room and Moor--2014 Multi-University Training Contest 6

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

Room and Moor

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 742    Accepted Submission(s): 220

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

 

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这道题，比赛时候没做出来，后面看了题解又想了好久才搞出来的。

对f(x)求导可以得到 在一段区间内，当x为平均值时，f(x)的值最小，但由于我们可以分成好多段，每一段的值可以是不同的。因此我们需要模拟一个栈来维护它。

从左到右进行扫描，遇到A[i],时把它加入栈内。然后检查栈顶是否有x[i]>x[i-1],若有则将其合并取平均值，为了方面我这边用num[i]表示第i个区间内1的数量，len[i]表示第i个段的长度。所以只需要判断 num[i]*len[i-1]>num[i-1]*len[i]的大小即可。

到最后，我们加上每个段的大小即可。需要注意的一点是num[i]统计的是第i段1的数量，那么len[i]-num[i] 统计的便是该段0的数量。将它们加起来即可。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<string>#include<vector>#include<map>#include<set>#include<cstdlib>#include<stack>#include<queue>#include<bitset>using namespace std;#define CLR(A) memset(A,0,sizeof(A))const int MAX=100010;double num[MAX];double len[MAX];int main(){    int T;    while(~scanf("%d",&T)){        while(T--){            int n,cnt=0;            CLR(num);CLR(len);            scanf("%d",&n);            for(int i=0;i<n;i++){                int v;scanf("%d",&v);                num[cnt]=v;                len[cnt]=1;                while(cnt>=1){                    if(num[cnt]*len[cnt-1]>num[cnt-1]*len[cnt]) break;                    num[cnt-1]+=num[cnt];                    len[cnt-1]+=len[cnt];                    cnt--;                }                cnt++;            }            double ret=0.0;            for(int i=0;i<cnt;i++){                double x=num[i]/len[i];                ret+=x*x*(len[i]-num[i])+(1-x)*(1-x)*num[i];            }            printf("%.6lf\n",ret);        }    }    return 0;}