opencv —— minMaxLoc 寻找图像全局最大最小值

#include<opencv.hpp>
#include<iostream>
#include<string>
using namespace std;
using namespace cv;
int main() {
    Mat src = imread("C:/Users/齐明洋/Desktop/证件照/6.jpg");
    imshow("src", src);
    
    //通道分离
    vector<Mat>images;
    split(src, images);
    
    //计算各通道直方图，并进行统计
    //https://www.cnblogs.com/bjxqmy/p/12378312.html
    vector<Mat>bgr(3);
    int channels[] = { 0,1,2 };
    int histsize[] = { 10,10,10 };
    float r[] = { 0,255 };
    const float *ranges[] = { r,r,r };
    Scalar colors[] = { Scalar(255,0,0),Scalar(0,255,0),Scalar(0,0,255) };
    string names[] = { "蓝色通道","绿色通道","红色通道" };
    for (int i = 0; i < 3; i++) {
        calcHist(&src, 3, &channels[i], Mat(), bgr[i], 1, &histsize[i], &ranges[i], true);
        normalize(bgr[i], bgr[i], 1, 0, NORM_L1);//https://www.cnblogs.com/bjxqmy/p/12292421.html
        double mx, mn; Point pmx, pmn;
        minMaxLoc(bgr[i], &mn, &mx, &pmn, &pmx);

        double width = 255.0 / histsize[i];
        cout << names[i] << endl;
        cout << "数量最少的颜色范围：[" << pmn.y*width << "," << (pmn.y + 1)*width << "] 占比：" << mn * 100 << "%" << endl;
        cout << "数量最多的颜色范围：[" << pmx.y*width << "," << (pmx.y + 1)*width << "] 占比：" << mx * 100 << "%" << endl;
        cout << endl;

        imshow(names[i], images[i]);
    }
    waitKey(0);
}