codeforces A. Cakeminator 題解

本題思路：

1 先掃描行，如果可以吃，就數吃了多少格，然後做好標志

2 掃描列，同樣處理

掃描完就可以出答案了。

時間效率是O(n*m)了。算是暴力法

題目：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

void Cakeminator()
{
	unsigned row, col;
	cin>>row>>col;
	vector cake(row);
	for (unsigned i = 0; i < row; i++)
	{
		cin>>cake[i];
	}
	int cakeCells = 0;
	for (unsigned i = 0; i < row; i++)
	{
		bool eatable = true;
		for (unsigned j = 0; j < col; j++)
		{
			if ('S' == cake[i][j]) eatable = false;
		}
		if (eatable)
		{
			for (unsigned j = 0; j < col; j++)
			{
				if ('.' == cake[i][j])
				{
					cake[i][j] = 'E';
					cakeCells++;
				}
			}
		}
	}
	for (unsigned j = 0; j < col; j++)
	{
		bool eatable = true;
		for (unsigned i = 0; i < row; i++)
		{
			if ('S' == cake[i][j]) eatable = false;
		}
		if (eatable)
		{
			for (unsigned i = 0; i < row; i++)
			{
				if ('.' == cake[i][j])
				{
					cake[i][j] = 'E';
					cakeCells++;
				}
			}
		}
	}
	cout<