PAT (Advanced Level) Practise 1054 The Dominant Color (20)

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本文介绍了一个算法挑战，任务是在给定的图像中找到占据超过一半面积的严格主导颜色。输入包括图像的分辨率和颜色矩阵，输出为主导颜色。文章提供了解决方案的思路及示例代码。

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1054. The Dominant Color (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 24

Sample Output:

题意：给你n*m个数，找出出现次数超过一半的是哪个

解题思路：遇到不同的元素，如果出现次数为0，更跟换成当前元素，如果次数不为0则-1，遇到相同元素，出现次数相加，最终记录的元素就是所求元素

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int main()
{
	int n, m;
	while (~scanf("%d%d", &m,&n))
	{
		int cnt=0, ans=0, x;
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < m; j++)
			{
				scanf("%d", &x);
				if (x == ans) cnt++;
				else
				{
					if (cnt) cnt--;
					else ans = x, cnt = 1;
				}
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}