【算法笔记】A1054 The Dominant Color
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:
1 #include<bits/stdc++.h> 2 using namespace std; 3 map<int, int> colorMap; 4 int main(){ 5 int m, n, color, domin = 0; 6 cin>>m>>n; 7 for(int i = 0; i < n; i++){ 8 for(int j = 0; j < m; j++){ 9 cin>>color; 10 colorMap[color]++; 11 if(colorMap[domin] < colorMap[color]) domin = color; 12 } 13 } 14 cout<<domin; 15 return 0; 16 }
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