題目來源： http://community.topcoder.com/stat?c=problem_statement&pm=3005&rd=5858

思路：

如果直接用Brute Force搜索所有可能的圓的話，那么搜索空間將很大，所以我用了一個priority_queue結(jié)構(gòu)，將搜索的順序變?yōu)榘磮A的半徑從大到小搜索，所以當(dāng)搜索到符合條件的圓時，即可停止搜索。這樣可以大大減少搜索范圍，不過對于最壞的情況，也就是沒有符合條件的圓時，還是會將所有的可能情況都搜索到。

代碼如下：

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    #include <iostream>

#include <algorithm>

#include <vector>

#include <queue>

#include <utility>

#include <cmath>



using namespace std;



typedef pair <int, pair<int, int> > entry;

#define make_entry(radius, row, col) ( make_pair( (radius), make_pair( (row), (col) ) ) )



class LargestCircle

{

public:

	int radius(vector <string> grid);

};



bool checkCircle(int cusRadius, int cus_row, int cus_col, vector<string> & grid);

int LargestCircle::radius(vector<string> grid)

{

	int i, j;

	int rows, cols, cus_row, cus_col;

	priority_queue <entry> PQ;

	int maxRadius, cusRadius;



	rows = grid.size();

	cols = grid[0].size();

	for (i = 0; i < rows - 1; i++) {

		for (j = 0; j < cols - 1; j++) {

			PQ.push(make_pair(min(min(i + 1, j + 1), min(rows - i - 1, cols - j - 1)), 

				make_pair(i, j)));

		}

	}



	maxRadius = 0;



	while ( !PQ.empty() ) {

		cusRadius = PQ.top().first;

		cus_row = PQ.top().second.first;

		cus_col = PQ.top().second.second;

		PQ.pop();



		if ( checkCircle(cusRadius, cus_row, cus_col, grid) ) {

			maxRadius = cusRadius;

			break;

		}

		--cusRadius;

		if (cusRadius > 0) {

			PQ.push(make_entry(cusRadius, cus_row, cus_col));

		}

	}



	return maxRadius;

}



/**

 * 檢查該圓是否合法

 */

bool checkCircle(int cusRadius, int cus_row, int cus_col, vector<string> & grid)

{

	int i, j;

	double pre_bias, next_bias;

	int pre_cell, next_cell;

	int next_row, next_col;



	next_bias = 0;

	for (i = 0; i < cusRadius; i++) {

		pre_bias = next_bias;	/* pre_bias值與上一個循環(huán)中的 next_bias值相等 */

		next_bias = sqrt( pow( (double)cusRadius, 2 ) - pow( (double)(cusRadius-i - 1), 2 ) );



		pre_cell = (int)pre_bias;

		next_cell = (int)next_bias;

		if ( abs( next_bias - next_cell > 0.00001)) {

			++next_cell;

		}



		for (j = 0; j < next_cell - pre_cell; j++) {

			next_row = cus_row - (cusRadius - i) + 1;

			next_col = cus_col - pre_cell - j;

			if ('#' == grid[next_row][next_col] ||

			    '#' == grid[next_row][2 * cus_col + 1 - next_col] ||	// 右對稱點(diǎn)

			    '#' == grid[2 * cus_row + 1 - next_row][next_col] ||	// 下對稱點(diǎn)

			    '#' == grid[2 * cus_row + 1 - next_row][2 * cus_col + 1 - next_col]) {	//中心對稱

				return false;

			}

		}

	}



	return true;

}


  

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Test SRM Level Three: LargestCircle, Brute Force