1054:The Dominant Color
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1054. The Dominant Color (20)
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.
Input Specification:
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.
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 24Sample Output:
24
思路
map模拟一个字典,把出现的颜色计数,然后遍历字典把出现次数最多的颜色打印出来就行。
注:
1.原本是用的unordered_map容器,但是有两个用例会超时,应该是搜索的时候数据太大超时了,换成map就AC掉了。
2.map内部结构是红黑树,每次插入会自动平衡树和对插入的键排序,unordered_map内部是哈希表。超时的测试用例数据应该是具有一定顺序性使得map的查找效率比unordered_map更高了。
代码
#include<iostream> #include<map> using namespace std; int main() { int N,M; while(cin >> N >> M) { map<int,int> dic; for(int i = 0;i < N;i++) { for(int j = 0;j < M;j++) { int value; cin >> value; if(dic.count(value) > 0) { dic[value]++; } else dic.insert(pair<int,int>(value,1)); } } pair<int,int> cur(0,0); for(map<int,int>::iterator it = dic.begin(); it != dic.end();it++) { if(it->second > cur.second) { cur.first = it->first; cur.second = it->second; } } cout << cur.first; } }
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