poj 1147 Binary codes

Binary codes

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5647		Accepted: 2201

Description

Consider a binary string (b₁…b_N) with N binary digits. Given such a string, the matrix of Figure 1 is formed from the rotated versions of the string.

b1	b2	…	bN−1	bN
b2	b3	…	bN	b1
…
bN−1	bN	…	bN−3	bN−2
bN	b1	…	bN−2	bN−1

Figure 1. The rotated matrix

Then rows of the matrix are sorted in alphabetical order, where ‘0’ is before ‘1’. You are to write a program which, given the last column of the sorted matrix, finds the first row of the sorted matrix.

As an example, consider the string (00110). The sorted matrix is

0	0	0	1	1
0	0	1	1	0
0	1	1	0	0
1	0	0	0	1
1	1	0	0	0

and the corresponding last column is (1 0 0 1 0). Given this last column your program should determine the first row, which is (0 0 0 1 1).

Input

The first line contains one integer N ≤ 3000, the number of binary digits in the binary string. The second line contains N integers, the binary digits in the last column from top to bottom.

Output

The first line contains N integers: the binary digits in the first row from left to right.

Sample Input

5
1 0 0 1 0

Sample Output

0 0 0 1 1

对由0,1组成的n个数。照题中的旋转，最后依据每行的字典序排序，组成n*n的矩阵，给出矩阵的最后1列，求矩阵

的第首行。

给出最后一列能够求出第0列，由于是按字典序排的，所以第0列肯定0在前，1在后。而第0列为0的相对位置在最后

1列不变。由于第0列都为0，又是按字典序排的。第0列为1也一样。依据第0列和最后一列就能够将相应关系求出。

也就是next数组。

比如例子的

0 0 0 1 1

0 0 1 1 0

0 1 1 0 0

1 0 0 0 1

1 1 0 0 0

next[ ]={1,2,4,0,3}

第0行第0列为0，第0行的第1列下次旋转后为第0列的第0行，所以第0行第1列为第0列的第1行为0，第1列的第0行

为第2列的第1行，为第0列的第2行。所以第0行的第2列为第0列的第2行为0，通过推理发现为next数组中元素递推

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=5000+100;
int last[maxn];
int first[maxn];
int next[maxn];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
           scanf("%d",&last[i]);
           first[i]=last[i];
        }
        sort(first,first+n);
        int cur=0;
        int i;
        for(i=0;i<n;i++)
        {
            if(first[i])
            break;
            while(last[cur]&&cur<n)
            cur++;
            next[i]=cur++;
        }
        cur=0;
        for(i=i;i<n;i++)
        {
            while(last[cur]==0&&cur<n)
            cur++;
            next[i]=cur++;
        }
        int  k=0;
        for(int i=0;i<n-1;i++)
        {
           printf("%d ",first[k]);
           k=next[k];
        }
        printf("%d
",first[k]);
    }
    return 0;
}