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