• Test SRM Level Three: LargestCircle, Brute Force


    题目来源:http://community.topcoder.com/stat?c=problem_statement&pm=3005&rd=5858


    思路:

    如果直接用Brute Force搜索所有可能的圆的话,那么搜索空间将很大,所以我用了一个priority_queue结构,将搜索的顺序变为按圆的半径从大到小搜索,所以当搜索到符合条件的圆时,即可停止搜索。这样可以大大减少搜索范围,不过对于最坏的情况,也就是没有符合条件的圆时,还是会将所有的可能情况都搜索到。


    代码如下:

    #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;
    }
    
    /**
     * 检查该圆是否合法
     */
    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值与上一个循环中的 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] ||	// 右对称点
    			    '#' == grid[2 * cus_row + 1 - next_row][next_col] ||	// 下对称点
    			    '#' == grid[2 * cus_row + 1 - next_row][2 * cus_col + 1 - next_col]) {	//中心对称
    				return false;
    			}
    		}
    	}
    
    	return true;
    }
    


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  • 原文地址:https://www.cnblogs.com/dyllove98/p/3170470.html
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