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,..., AN}, each of which is either 0 or 1. In order to beat him, programmer Moor has to construct another sequence B = {B1, B2,... , BN} 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 Ai.
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 Ai.
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
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
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);
}
}