The Dominant Color (20)

最新推荐文章于 2022-11-16 22:18:30 发布

原创最新推荐文章于 2022-11-16 22:18:30 发布 · 400 阅读

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本文介绍了一个算法挑战，旨在从给定分辨率的图像中找出占据面积超过一半的严格主导颜色。提供了用C、C++及Java实现的示例代码。

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题目描述

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.

输入描述:

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.

输出描述:

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

输入例子:

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

输出例子:

我的代码（C语言）：

#include<stdio.h>
int n,m,i,j,x=0,y,a[1001][1001];
int main()
{
    scanf("%d%d",&n,&m);
    for(i=0;i<m;i++)
    {
        for(j=0;j<n;j++)
        {
            scanf("%d",&a[i][j]);
            if(x==0)
            {
                y=a[i][j];
                x++;
            }
            else
            {
                if(y!=a[i][j]) x--;
                else x++;
            }
        }
    }
    printf("%d\n",y);
    return 0;
}

我的代码（C++）：

#include<iostream>
using namespace std;
int n,m,i,j,x=0,y,a[1001][1001];
int main()
{
    cin>>n>>m;
    for(i=0;i<m;i++)
    {
        for(j=0;j<n;j++)
        {
            cin>>a[i][j];
            if(x==0)
            {
                y=a[i][j];
                x++;
            }
            else
            {
                if(y!=a[i][j]) x--;
                else x++;
            }
        }
    }
    cout<<y<<endl;
    return 0;
}

我的代码（Java）：

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner In = new Scanner(System.in);
        int x=0,y=0,i,j,a[][]=new int [1001][1001];
        int n=In.nextInt();
        int m=In.nextInt();
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                a[i][j]=In.nextInt();
                if(x==0)
                {
                    y=a[i][j];
                    x++;
                }
                else
                {
                    if(y!=a[i][j]) x--;
                    else x++;
                }
            }
        }
        System.out.printf("%d\n",y);
    }
}