1054 The Dominant Color (20分)-CFANZ编程社区

文章目录

1 题目
2 解析

2.1 题意
2.2 思路

2.2.1 思路一（map数组）
2.2.2 思路二（不用map数组）

3 参考代码

3.1思路一
3.2 思路二

1 题目

1054 The Dominant Color (20分)
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, 800×600), 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 [0,2
24
). 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

2 解析

2.1 题意

统计矩阵块中，最占优势的颜色（出现次数最多的颜色）

2.2 思路

2.2.1 思路一（map数组）

如果直接开一个int类型hash数组，由于数字范围最大可达16777216，可能会导致内存空间超限，因此可以使用map<int,int>进行映射，对于每次输入的值用find函数查找是否出现过，如果出现过则加一，否则将值赋值为1。

2.2.2 思路二（不用map数组）

由于题目最优颜色的数量要求超过半数，因此有超过半数的数相同，如果采用两两不同相互抵消的做法。

设置一个ans记录答案，count统计ans出现的次数，
然后在读入时，判断ans与读入的数字是否相同，如果不想等，则抵消一次（即令count减1），如果相等，则令count加1。如果某步count被抵消到0，令新的数字为ans。最后剩下的ans就是答案。

3 参考代码

3.1思路一

#include 
#include 

using std::map;

map<int , int> count;

int main(int argc, char const *argv[])
{
    int m, n;
    scanf("%d%d", &m , &n);
    

    int num;
    for (int i = 0; i < n; ++i)
    {
        for (int j = 0; j < m; ++j)
        {
            scanf("%d", &num);
            if(count.find(num) != count.end()){
                count[num]++;
            }else{
                count[num] = 1;
            }
        }
    }

    int max = -1, idx = 0;
    for (map<int,int>::iterator it = count.begin(); it != count.end(); ++it)
    {
            if(it->second > max){
                max = it->second;
                idx = it->first;
            }
    }

    printf("%d\n", idx);

    return 0;
}

3.2 思路二

/*
3 2
4502761 4502761
4502761 4502761
7340438 3208719
 */

#include 

int main(int argc, char const *argv[])
{
    int n, m;
    scanf("%d%d", &n, &m);

    int num;
    int count = 1, ans = -1;
    for(int i  = 0; i < n * m; i++){
        scanf("%d", &num);
        if(ans != num){
            count--;
        }else{
            count++;
        }
        if(count == 0){
            ans = num;
            count++;
        }


    }
    printf("%d\n", ans);
    return 0;
}