【算法笔记】A1054 The Dominant Color

1054 The Dominant Color （20 分)

 

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, 8), 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 (≤) and N (≤) which are the resolutions of the image. Then N lines follow, each contains M digital colors in the range [0). 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:

24

题意

　　给出一张图片的分辨率，在下面的n行m列中给出每个像素点的颜色编码，最后输出数量最多的颜色的编码。

思路：

　　用map记录每个颜色出现的次数，很容易实现输出出现次数最多颜色。

　　第一次失误写成了 cout<<color; 竟然AC了。。。难道每个测试用例最后输入的颜色就是答案？把map去掉只留下输入color输出color，还真的全部AC。可惜输入m*n个数时间复杂度还是O(m*n)，跟原来的耗时没什么区别，暂时还没什么方法把时间复杂度降到O(1)。

code：

#include<bits/stdc++.h>
using namespace std;
map<int, int> colorMap;
int main(){
    int m, n, color, domin = 0;
    cin>>m>>n;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            cin>>color;
            colorMap[color]++;
            if(colorMap[domin] < colorMap[color]) domin = color;
        }
    }
    cout<<domin;
    return 0;
}