PAT 1054 The Dominant Color[简单][运行超时的问题]
1054 The Dominant Color (20)(20 分)
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 strictlydominant 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, 2^24^). 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:计算矩阵中出现次数超过半数的数字,数字范围是1-2^24.
//一开始我的提交是这样的,牛客网上可以通过,但是pat上运行测试点3超时。
#include <iostream>
#include<stdio.h>
#include<map>
using namespace std;
map<int,int> mp;
int main() {
int m,n,pix;
int mx=,mpx;
cin>>m>>n;
for(int i=;i<n;i++)
for(int j=;j<m;j++){
cin>>pix;
if(mp[pix]==)
mp[pix]=;
else
mp[pix]++;
if(mp[pix]>mx){
mpx=pix;
mx=mp[pix];
}
}
cout<<mpx;
return ;
}
1.参考了大佬的代码,发现大佬是直接判断如果出现超过半数,那么就直接Return 0;
2.修改return 0;之后提交仍是超时,就将mp==0去掉,直接出现就mp[pix]++;
3.提交之后还是超时,以为是数据的问题,改为map<long,int>还是不行;
4.最终将cin改为scanf即可,数据量大的时候cin读入会超时啊!
最终AC代码如下:
#include <iostream>
#include<stdio.h>
#include<map>
using namespace std;
map<int,int> mp;
int main() {
int m,n;
int pix;
scanf("%d %d",&m,&n);
int len=m*n/;
for(int i=;i<n;i++)
for(int j=;j<m;j++){
scanf("%d",&pix);
mp[pix]++;
if(mp[pix]>len){
printf("%d",pix);
return ;
}
} return ;
}
//第一次体会到,cin真的会超时。
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