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 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;
}