pat1054. The Dominant Color (20)

1054. The Dominant Color (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

---

提交代码

方法一：

#include <cstdio>
int main(){
    int m, n, x, times = 0, candidate;
    scanf("%d%d", &m, &n);
    for (int i = 0; i < n; ++i){
        for (int j = 0; j < m; ++j){
            scanf("%d", &x);
            if (times == 0){
                candidate = x;
                ++times;
            }
            else{
                if (x == candidate)
                    ++times;
                else
                    --times;
            }
        }
    }
    printf("%d
", candidate);
    return 0;
}

方法二：map

#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
map<string,int> img;
int main(){
    //freopen("D:\INPUT.txt","r",stdin);
    int m,n,i,j;
    scanf("%d %d",&m,&n);
    char color[15];
    for(i=0;i<m;i++){
        for(j=0;j<n;j++){
            scanf("%s",color);
            img[color]++;
        }
    }
    map<string,int>::iterator it;
    int half=m*n/2+1;
    for(it=img.begin();it!=img.end();it++){
        if(it->second>=half){
            break;
        }
    }
    cout<<it->first<<endl;
    return 0;
}