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最新推荐文章于 2025-06-18 17:49:18 发布

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本文探讨如何通过标记统计法解决图像处理中确定主导颜色的问题。通过输入图像的宽度和高度，以及图像中各像素的颜色值，算法能找出占据图像大部分区域的主导颜色。

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http://pat.zju.edu.cn/contests/pat-a-practise/1054

1054. The Dominant Color (20)

时间限制

100 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

Behind the scenes in the computer's memory, color is always talked about as a series of 24 bits of information for each pixel. In an image, the color with the largest proportional area is called the dominant color. A strictly dominant color takes more than half of the total area. Now given an image of resolution M by N (for example, 800x600), you are supposed to point out the strictly dominant color.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: M (<=800) and N (<=600) which are the resolutions of the image. Then N lines follow, each contains M digital colors in the range [0, 2²⁴). It is guaranteed that the strictly dominant color exists for each input image. All the numbers in a line are separated by a space.

Output Specification:

For each test case, simply print the dominant color in a line.

Sample Input:

5 3
0 0 255 16777215 24
24 24 0 0 24
24 0 24 24 24

Sample Output:

思路：标记统计法，因为肯定会有一个数的个数超过总数的一半，那么记下之前读到的数的个数cnt和数字last，当前读到的和last相同时，++cnt，而当当前读到的与last不同时，分两种情况，当cnt=1，cnt不变并直接把当前的赋给last，而当cnt>1时，--cnt并且last保持不变。最后直接输出last。这其实就是类似于“连连看”，只不过是消除的是两个不一样的元素，那么剩下的肯定就是占大多数的元素。

参考代码：

#include<cstdio>

int main()
{
    int M, N, num, last, cnt=0;
    scanf("%d %d", &M, &N);
    for(int i = 0; i < M; ++i)
    {
        for(int j = 0; j < N; ++j)
        {
            scanf("%d", &num);
            if(i == 0 && j==0)
            {
                cnt = 1;
                last = num;
            }
            else
            {
                if(last == num)
                {
                    ++cnt;
                }
                else
                {
                    if(cnt > 1)
                    {
                        --cnt;
                    }
                    else
                    {
                        last = num;
                    }
                }
            }
        }
    }
    printf("%d\n", last);
    return 0;
}