PAT甲级——【牛客A1005】

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:

24

版本一:

开始我以为主导颜色是指要连成一片的才算，零散分布不算，所以想了一个关于“岛问题”的解决方法，见版本二，后来才发现，原来就是数数，谁多，谁就是主导色

 #include <iostream>
 #include <vector>
 #include <map>

 using namespace std;

 int main()
 {
     int M,N;
     int maxColor = ;
     int resColor = ;
     cin >> M >> N;
     map<int, int>res;
     vector<vector<int>>data(M, vector<int>(N, ));
     for (int i = ; i < N; ++i)
     {
         for (int j = ; j < M; ++j)
         {
             int a;
             cin >> a;
             res[a]++;
         }
     }
     for (auto ptr = res.begin(); ptr != res.end(); ++ptr)
     {
         if(ptr->second > maxColor)
         {
             maxColor = ptr->second;
             resColor = ptr->first;
         }
     }
     cout << resColor << endl;
     return ;

 }

例版本一：

关于“岛问题”，使用递归遍历，就是对于每一种颜色，通过上下左右遍历出其所有的相同的颜色，并计数和标记，则得到每种颜色最多的主导色

#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;

int M, N;

//方法一，使用

void Travle(vector<vector<int>>&data, int a, int b, int &num, const int color)
{
    if ( a <  || b <  || a >= M || b >= N|| data[a][b] != color)
        return;
    num++;
    data[a][b] = -;//标记已遍历
    Travle(data, a - , b, num, color);
    Travle(data, a + , b, num, color);
    Travle(data, a, b - , num, color);
    Travle(data, a, b + , num, color);
}

int main()
{
    Test1();

    //M*N
    int maxColor = ;
    int resColor = ;
    cin >> M >> N;
    vector<vector<int>>data(M, vector<int>(N, ));
    for (int i = ; i < N; ++i)
        for (int j = ; j < M; ++j)
            cin >> data[j][i];

    for (int i = ; i < M; ++i)
    {
        for (int j = ; j < N; ++j)
        {
            if (data[i][j] >= )//未遍历过
            {
                int num = ;
                int color = data[i][j];
                Travle(data, i, j, num, color);
                if (num > maxColor)
                {
                    maxColor = num;
                    resColor = color;
                }
            }
        }
    }

    cout << resColor << endl;
    return ;

}

版本三：

　　笨死了，英语理解能力不好，其实就是监测每次的输入，当输入某种颜色的个数过半，则立马输出!

 #include <iostream>
 #include <vector>
 #include <unordered_map>

 using namespace std;

 int main()
 {
    int M, N;
     cin >> M >> N;
     unordered_map<int, int>res;
     vector<vector<int>>data(M, vector<int>(N, ));
     for (int i = ; i < N; ++i)
     {
         for (int j = ; j < M; ++j)
         {
             int a;
             cin >> a;
             res[a]++;
             if (res[a] > M*N / )
             {
                 cout << a << endl;
                 break;
             }
         }
     }
     return ;

 }