【PAT】1054 The Diamond Color

题目：http://pat.zju.edu.cn/contests/pat-a-practise/1054

题目描述：

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. Astrictly 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

普通的思路:标记每一个数出现的次数，结果会超时，有一组数据过不去。

#include<iostream>
using namespace std;

#define max 170000000
int input[max];
long t,temp;
int main()
{
	int M,N;
	int count = 0;
	int i,j;
	cin>>M>>N;
 
	//采用以下for循环会超时
	for(j=0; j<M*N; j++)
	{
		cin>>t;
		input[t]++;
		if(input[t] > count) {count = input[t]; temp = t;}
	}
 
	cout<<temp<<endl;	
	return 0;
}

正确解法一：采用map。

代码来自于：http://blog.csdn.net/matrix_reloaded/article/details/8818628

#pragma warning (disable:4786)
#include<iostream>
#include<string>
#include<cstdio>
#include<map>
using namespace std;

int main()
{
	int m,n;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		int i,j;
		string s;
		map<string,int> ans;
		
		map<string,int>::iterator p;

		char ss[10000];

		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{
				scanf("%s",ss);
				s=ss;
				ans[s]++;
			}
		}
		int cnt=-1;
		for(p=ans.begin();p!=ans.end();p++)
		{
			if(p->second>cnt)
			{
				s=p->first;
				cnt=p->second;
			}
		}
		printf("%s\n",s.c_str());
	}
	return 0;
}

正确解法二：因为我们所要求的数出现的次数超过总数目一半，所以将不相同的数两两抵消，那么最终剩下的便是我们要求的数。

代码来自于：http://blog.csdn.net/jjike/article/details/8970540

#include<iostream>
using namespace std;

const int MAXN = 10000;
int img[MAXN]={0};

int main()
{
	int m,n,mn;
	int i,j,x;
	freopen("C:\\Documents and Settings\\Administrator\\桌面\\input.txt","r",stdin);

	bool fg = false;
	//cin>>m>>n;
	scanf("%d%d",&m,&n);
	mn = m * n;
	for(i = 0; i < mn; i++){		
		//cin>>img[i];
		scanf("%d",&img[i]);		
	}
	int cad, nTime=0;
	for(i = 0; i < mn; i++){
		if(nTime == 0){
			cad = img[i];
			nTime = 1;
		} else {
			if(cad == img[i])
				nTime++;
			else 
				nTime--;			
		}
	}
	//cout<<cad;
	printf("%d",cad);
	return 0;
}

也可以简单些，用vector就好：

#include<iostream>
#include<vector>
using namespace std;
int main(){
	int n,k;
	cin>>n>>k;
	int i;
	vector<long> vec;
	long t;
	for(i=0; i<n*k; i++){
		scanf("%ld",&t);
		if(vec.empty() || vec.front() == t ){
			vec.push_back(t);
		}
		else
			vec.pop_back();
	}
	cout<<vec.front()<<endl;
	return 0;
}