【22.70%】【codeforces 591C】 Median Smoothing

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia article that described the method of median smoothing (or median filter) and its many applications in science and engineering. Vasya liked the idea of the method very much, and he decided to try it in practice.

Applying the simplest variant of median smoothing to the sequence of numbers a1, a2, …, an will result a new sequence b1, b2, …, bn obtained by the following algorithm:

b1 = a1, bn = an, that is, the first and the last number of the new sequence match the corresponding numbers of the original sequence.
For i = 2, …, n - 1 value bi is equal to the median of three values ai - 1, ai and ai + 1.
The median of a set of three numbers is the number that goes on the second place, when these three numbers are written in the non-decreasing order. For example, the median of the set 5, 1, 2 is number 2, and the median of set 1, 0, 1 is equal to 1.

In order to make the task easier, Vasya decided to apply the method to sequences consisting of zeros and ones only.

Having made the procedure once, Vasya looked at the resulting sequence and thought: what if I apply the algorithm to it once again, and then apply it to the next result, and so on? Vasya tried a couple of examples and found out that after some number of median smoothing algorithm applications the sequence can stop changing. We say that the sequence is stable, if it does not change when the median smoothing is applied to it.

Now Vasya wonders, whether the sequence always eventually becomes stable. He asks you to write a program that, given a sequence of zeros and ones, will determine whether it ever becomes stable. Moreover, if it ever becomes stable, then you should determine what will it look like and how many times one needs to apply the median smoothing algorithm to initial sequence in order to obtain a stable one.

Input
The first input line of the input contains a single integer n (3 ≤ n ≤ 500 000) — the length of the initial sequence.

The next line contains n integers a1, a2, …, an (ai = 0 or ai = 1), giving the initial sequence itself.

Output
If the sequence will never become stable, print a single number  - 1.

Otherwise, first print a single integer — the minimum number of times one needs to apply the median smoothing algorithm to the initial sequence before it becomes is stable. In the second line print n numbers separated by a space — the resulting sequence itself.

Examples
input
4
0 0 1 1
output
0
0 0 1 1
input
5
0 1 0 1 0
output
2
0 0 0 0 0
Note
In the second sample the stabilization occurs in two steps: , and the sequence 00000 is obviously stable.

【题目链接】:http://codeforces.com/contest/591/problem/C

【题解】

按照题目的要求；
只要出现了连续的1和连续的0(连续的个数大于等于2);
则这些连续的数字肯定不会再发生变化了;
然后再考虑那些01交替出现的情况;
比如0000010101010111111
中间的10101010是交替出现的,这些都会发生变化;
显然变一次会变成
01010101即全部取反;
则最左边的0和最右边的1会和原本这个01串两边的连续串“融合”在一起；
这样就缩小了规模
变成了
0000001010101111111
然后会发生变化的就变成了
101010
再变
010101(取反)
则最左边和最优边分别又有一个0和1和边界“融合”了；
就这样i-j/2次之后显然就不会再发生变化了；
而最左边和最右边会相应的变成这个01串原本的左边连续串和右边连续串的值;
然后在所有的01串中取相应的(i-j)/2的max值即可,即为答案;

【完整代码】

#include <cstdio>#include <cstdlib>#include <cmath>#include <set>#include <map>#include <iostream>#include <algorithm>#include <cstring>#include <queue>#include <vector>#include <stack>#include <string>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long longusing namespace std;const int MAXN = 509999;const int dx[5] = {0,1,-1,0,0};const int dy[5] = {0,0,0,-1,1};const double pi = acos(-1.0);int n;int a[MAXN];void rel(LL &r){    r = 0;    char t = getchar();    while (!isdigit(t) && t!='-') t = getchar();    LL sign = 1;    if (t == '-')sign = -1;    while (!isdigit(t)) t = getchar();    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();    r = r*sign;}void rei(int &r){    r = 0;    char t = getchar();    while (!isdigit(t)&&t!='-') t = getchar();    int sign = 1;    if (t == '-')sign = -1;    while (!isdigit(t)) t = getchar();    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();    r = r*sign;}int main(){    //freopen("F:\\rush.txt","r",stdin);    rei(n);    for (int i = 1;i <= n;i++)        rei(a[i]);    int m = 0;    for (int i = 1;i <=n-1;i++)        if (a[i]!=a[i+1])        {            int j = i+1;            while (j+1<=n && a[j+1]!=a[j]) j++;            m = max((j-i)/2,m);            int l = i+1,r = j-1;            while (l <= r)            {                a[l] = a[i];                a[r] = a[j];                l++;r--;            }            i = j;        }    cout << m<<endl;    for (int i = 1;i <= n;i++)        printf("%d ",a[i]);    return 0;}