hdu Median Filter

Median Filter

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 74    Accepted Submission(s): 31

Problem Description

Median filter is a cornerstone of modern image processing and is used extensively in
Given a black and white image of n by n pixels, the algorithm works as follow: For each pixel p at the i‐th row and the j‐th column (1+ r <= i, j <= n – r), its gray level g[i,j] is replace by the median of gray levels of pixels in the (2r+1) by (2r+1) square centered at p. The square is called the filtering window and r is its radius.
Considering the above example, the gray level of the pixel at center will be changed from 150 to 124, which is the median of a filtering window of radius 1.
Note that the algorithm won’t be applied on the pixels near the boundary, for the filtering window lies outside the image. So you are actually asked to output the filtered sub-image which contains the pixels from (r+1, r+1) to (n-r, n-r).

 

Input

The input contains several test cases. For each test case:
  (a) The first line contains two integers, n and r, meaning that the size of image is n * n (3 <= n <= 500) and the radius of filtering window is r ( 3 <= 2r + 1 <= n).
  (b) The following n lines contains the n by n gray level matrix presenting the image
  (c) The gray level ranges from 0 to 1000000 The input ends by double 0s.

 

Output

            For each test case output a (n – 2r) by (n – 2r) matrix presenting the sub-image after filtered.

 

Sample Input

3 1
1 1 1
1 1 1
1 1 1
3 1
1 9 6
4 5 2
3 7 8
3 1
0 0 0
255 255 255
0 255 0
5 1
0 0 1 1 0
1 0 1 0 1
0 0 1 1 1
1 1 1 0 1
1 0 0 0 1
0 0

 

Sample Output

1
5
0
0 1 1
1 1 1
1 0 1

Hint

The definition of “median”: One type of average, found by arranging the values in order and then selecting the one in the middle. If the total number of values in the sample is even, then the median is the mean of the two middle numbers. The median is a useful number in cases where the distribution has very large extreme values which would otherwise skew the data.

 

Source

The 35th ACM/ICPC Asia Regional Hangzhou Site —— Online Contest

 

Recommend

lcy

 

分析：树状数组求第k大的数，采用S型路线节约时间，另外调用很多的函数要写成inline

#include<cstdio>
#include<cstring>
#define maxn 1000002
int s[maxn];
int map[505][505];
int ans[505][505];
int MAX;

inline int lowbit(int x) {
    return x & (-x);
}

int sum(int a) {
    int ans = 0;
    while (a > 0) {
        ans += s[a];
        a -= lowbit(a);
    }
    return ans;
}

void add(int a, int c) {
    while (a <= MAX) {
        s[a] += c;
        a += lowbit(a);
    }
}

int find_kth(int k) {
    int sum = 0, pos = 0, i;
    for (i = 20; i >= 0; i--) {
        pos += (1 << i);
        if (pos > MAX || sum + s[pos] >= k) {
            pos -= (1 << i);
        } else {
            sum += s[pos];
        }
    }
    return pos + 1;
}

int main() {
    int n, i, j, r, k, R, t;
    while (scanf("%d%d", &n, &r) != EOF && n) {
        MAX = -1;
        memset(s, 0, sizeof (s));
        for (i = 1; i <= n; ++i)
            for (j = 1; j <= n; ++j) {
                scanf("%d", &map[i][j]);
                map[i][j]++;
                if (map[i][j] > MAX)
                    MAX = map[i][j];
            }
        R = 2 * r + 1;
        k = R * R / 2 + 1;
        for (i = 1; i <= R; ++i) {
            for (j = 1; j <= R; ++j) {
                add(map[i][j], 1);
            }
        }
        for (i = 1+r; i <= n -r; ++i) {
            if ((i-r) & 1) {
                if (i != 1+r) {
                    for (j = 1; j <= R; ++j) {
                        add(map[i -r- 1][j], -1);
                        add(map[i +r][j], 1);
                    }
                }
                ans[i][1 + r] = find_kth(k);
                for (j = 2+r; j <= n - r; ++j) {
                    for (t = i-r; t <= i + r; ++t) {
                        add(map[t][j - r-1], -1);
                        add(map[t][j + r], 1);
                    }
                    ans[i][j] = find_kth(k);
                }
            } else {
                for (j = n; j > n - R; --j) {
                    add(map[i - r-1][j], -1);
                    add(map[i + r][j], 1);
                }
                ans[i][n - r] = find_kth(k);
                for (j = n - r-1; j >= 1+r; --j) {
                    for (t = i-r; t <= i + r; ++t) {
                        add(map[t][j+r+1], -1);
                        add(map[t][j-r], 1);
                    }
                    ans[i][j] = find_kth(k);
                }
            }
        }
        for (i = 1 + r; i <= n - r; ++i) {
            for (j = 1 + r; j < n - r; ++j)
                printf("%d ", ans[i][j] - 1);
            printf("%d \n", ans[i][j] - 1);
        }
    }
    return 0;
}