PAT 1054. The Dominant Color (20)（简单排序题（`vector<int>`利用`map<int,int>`来排序））

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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

题目大意

• 1.给出一个二维数组，求出存放在二维数组里面那个数出现的频率最高。

解题思路

• 1.我这里用map<int,int>来储存每个数出现的次数，用vector<int>来储存这些出现了的数，最后vector<int>利用map<int,int>来排序。

AC代码

#include<vector>#include<iostream>#include<cstdio>#include<algorithm>#include<map>using namespace std;int n,m;map<int,int> wap;vector<int> keep;//`vector<int>`利用`map<int,int>`来排序bool cmp(int a,int b){    return wap[a] > wap[b];}int main(int argc, char *argv[]){    //scanf("%d %d",&n,&m);    cin >> n >> m;    int tem;    for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j) {            scanf("%d",&tem);            //用`map<int,int>`来储存每个数出现的次数            if (wap.find(tem)==wap.end()) {                wap[tem] = 0;                keep.push_back(tem);            }else {                wap[tem]++;            }        }    }    //`vector<int>`利用`map<int,int>`来排序    sort(keep.begin(),keep.end(),cmp);    cout << keep[0] << endl;    return 0;}