n*n的棋盘摆K的棋子,任意两个棋子不能在同一行同一列 Sample Input 2 1#. //# 可放.#4 4...#..#..#..#...-1 -1Sample Output 21 # include <cstdio> # include <cstring> # include <iostream> using namespace std ; ][] ; ] ; int sum ; int n , k ; void DFS(int src , int num
前言 之前一直想不明白dfs的时间复杂度是怎么算的,前几天想了下大概想明白了,现在记录一下. 存图方式都是链式前向星或邻接矩阵.主要通过几道经典题目来阐述dfs时间复杂度的计算方法. $n$是图中结点的个数,$e$是图中边的个数. 深度优先遍历图的每一个结点 时间复杂度为:链式前向星:$O\left( n + e \right)$:邻接矩阵:$O\left( n \right)$ 给定我们一个图(链式前向星存储),通过深度优先遍历的方式把图中每个结点遍历一遍. 首先,图中每个结点最多被遍历一次,
http://codeforces.com/problemset/problem/259/A PS9.8日做了但是忘了发博客,所以坚持3天了呦~ 终于读懂了这个题了,心累 Describe: 给你8 * 8的棋盘摆放问题,行的顺序可能是错乱的,问给你的8行是否能组成棋盘 Solution: 所以我们要检查的就是 1.棋盘有没有相邻的颜色相同 2.开头必须得是4 白 + 4黑(嘿嘿嘿,一开始我就是这么想的,但是!!!题目中右For that the friends can choose any r
枚举类型 枚举类型在JDK 5时引入. enum WeekEnum { MONDAY, TUESDAY, WEDNESDAY, THURDAY, FRIDAY, SATURDAY, SUNDAY } Java的枚举类java.lang.Enum中,有: |--private final String name;(定义时可表意的标识符) |--private final int ordinal;(序数,从0开始) |--及对应的获取方法. public class Enum1Common { en
Description Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had to use his toilet paper to draw on, for all of his paper was filled with squares and r