1054. The Dominant Color (20)

简介: Behind the scenes in the computer's memory, color is always talked about as a series of 24 bits of information for each pixel.

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:

24

常规思维,超时算法:

#include <iostream>
using namespace std;

int main(int argc, const char * argv[]) {
    int m, n;
    cin >> m >> n;
    string s[m*n], t;
    int num = 0, nc[480000] = {0};
    
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            int flag = 1;
            cin >> t;
            for (int k = 0; k < num; k++) {
                if(t == s[k]){
                    flag = 0;
                    nc[k]++;
                }
            }
            if (flag) {
                s[num] = t;
                nc[num++] = 1;
            }
        }
    }
    
    for (int i = 0; i < num; i++) {
        if (nc[i] > m*n / 2) {
            cout << s[i] << endl;
        }
    }
    return 0;
}
使用map的算法:
#include <cstdio>
#include <map>
using namespace std;
int main() {
    int m, n;
    scanf("%d %d", &m, &n);
    map<int, int> arr;
    int half = m * n / 2;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            int temp;
            scanf("%d", &temp);
            arr[temp]++;
            if(arr[temp] > half) {
                printf("%d", temp);
                return 0;
            }
        }
    }
    return 0;
}



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