PAT 1054 The Dominant Color

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 [0,224). 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

The question ： Give a 5*3 Matrix , Find the number in the matrix that is more than half of the number in the matrix and appears the most times , namely 24;

Ideas ：

Because the input data (16777215,1*10^7) Greater than int Maximum range （ The maximum range of integers is 8388608, namely 8*10^6）, So you can't use a hash table of arrays here , Should use the map As hash table , stay map of use find function , If you can find it, you can count ++, If it cannot be found, it means that this number does not exist before , Set as 1, Last traversal map, Just find the output with the most times

Study ：

1.map There is no need to open the size range for the definition of

2.map The traversal ,for(map< type , type >::iterator it=map.begin();it!=map.end();it++)

3.map Of key and value The interview of ,key:it->first,value:it->second;

Code ：

//map Use ;
// The title says that the maximum range of input numbers is to 2^24, Already exceeded int The maximum range of , Opening a hash table so large may cause hyperspace , So here is map; 
#include<iostream>
#include<map>
using namespace std;
int main(){
	int n,m,num;
	scanf("%d %d",&n,&m);
	map<int ,int > count;
	for(int i=0;i<n;i++){// Don't use it directly n*m, Multiplication of two integer numbers is easy to overflow ; 
		for(int j=0;j<m;j++){
			scanf("%d",&num); 
			if(count.find(num)!=count.end()){// If present, count , If not, create and set to 1; 
				count[num]++;
			}else {
				count[num]=1;
			}
		}
	}
	int max=-1;
	for(map<int,int>::iterator it=count.begin();it!=count.end();it++){// Iterative way 
	//for( type ::iterator it= type .begin(),it!= type .end();it++) 
		if(it->second>max){
			max=it->second;//it It can represent map An item of ; 
			num=it->first;
		}
	}
	printf("%d",num);
	return 0;
}