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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 24Sample Output:
24
方法一：
1 #include <cstdio> 2 int main(){ 3 int m, n, x, times = 0, candidate; 4 scanf("%d%d", &m, &n); 5 for (int i = 0; i < n; ++i){ 6 for (int j = 0; j < m; ++j){ 7 scanf("%d", &x); 8 if (times == 0){ 9 candidate = x; 10 ++times; 11 } 12 else{ 13 if (x == candidate) 14 ++times; 15 else 16 --times; 17 } 18 } 19 } 20 printf("%d\n", candidate); 21 return 0; 22 }
方法二：map
1 #include <iostream> 2 #include <cstdio> 3 #include <map> 4 using namespace std; 5 map<string,int> img; 6 int main(){ 7 //freopen("D:\\INPUT.txt","r",stdin); 8 int m,n,i,j; 9 scanf("%d %d",&m,&n); 10 char color[15]; 11 for(i=0;i<m;i++){ 12 for(j=0;j<n;j++){ 13 scanf("%s",color); 14 img[color]++; 15 } 16 } 17 map<string,int>::iterator it; 18 int half=m*n/2+1; 19 for(it=img.begin();it!=img.end();it++){ 20 if(it->second>=half){ 21 break; 22 } 23 } 24 cout<<it->first<<endl; 25 return 0; 26 }
pat1054. The Dominant Color (20)
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原文地址：http://www.cnblogs.com/Deribs4/p/4773026.html
