Balloons(DFS)】的更多相关文章

Link 题意: 给出一棵树,要求为其染色,并且使任意节点都不与距离2以下的节点颜色相同 思路: 直接DFS.由某节点出发的DFS序列,对于其个儿子的cnt数+1,那么因为DFS遍历的性质可保证兄弟结点的颜色不同,只需考虑当前节点是否与父亲结点和父亲的父亲结点颜色是否相同. /** @Date : 2017-05-09 22:37:25 * @FileName: 782C DFS.cpp * @Platform: Windows * @Author : Lweleth (SoundEarlf@g…
http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2248 题意:(1)求图中四连块(有公共边的方块)的个数: (2)求图中八连块(有公共顶点的方块)的个数. #include<stdio.h> #include<string.h> ; int vis1[N][N],map[N][N],vis[N][N]; void dfs1(int x,int y)//搜索四个方向 { if (!…
Balloons Time Limit: 1000MS Memory limit: 65536K 题目描述 Both Saya and Kudo like balloons. One day, they heard that in the central park, there will be thousands of people fly balloons to pattern a big image.They were very interested about this event, an…
Andryusha goes through a park each day. The squares and paths between them look boring to Andryusha, so he decided to decorate them. The park consists of n squares connected with (n - 1) bidirectional paths in such a way that any square is reachable…
从任意点出发,贪心染色即可. #include<cstdio> #include<algorithm> using namespace std; int v[200010<<1],next[200010<<1],first[200010],e; void AddEdge(int U,int V) { v[++e]=V; next[e]=first[U]; first[U]=e; } bool vis[200010]; int n,col[200010]; v…
http://codeforces.com/contest/782/problem/C 题意:给一棵树染最少的颜色,使得相邻距离为2的点都是不同的颜色,问最少是多少种颜色并输出每个点的颜色. 思路:比赛的时候没想到是找度最大的一个点并+1就是总颜色数,一直想怎么构造. 最后的总颜色数是度最大的一个点的度数+1.因为我们选的这个点到儿子的距离为1,因此其某一个儿子到另一个儿子的距离为2,就是这些儿子都要染成不同的颜色,+1是因为自己本身也要是不同的颜色. 然后确定了总颜色数,就可以DFS染色了.…
Crashing Balloon Time Limit: 2 Seconds      Memory Limit: 65536 KB On every June 1st, the Children's Day, there will be a game named "crashing balloon" on TV.   The rule is very simple.  On the ground there are 100 labeled balloons, with the num…
Balloons Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述 Both Saya and Kudo like balloons. One day, they heard that in the central park, there will be thousands of people fly balloons to pattern a big image. They were very interested about…
http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2152 Balloons Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述 Both Saya and Kudo like balloons. One day, they heard that in the central park, there will be thousands of…
C. Andryusha and Colored Balloons time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Andryusha goes through a park each day. The squares and paths between them look boring to Andryusha, so he…
C. Andryusha and Colored Balloons time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Andryusha goes through a park each day. The squares and paths between them look boring to Andryusha, so he…
2474 - Balloons in a Box 题目连接: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=99999999&page=show_problem&category=36&problem=475&mosmsg=Submission+received+with+ID+1967446 Description You must write a progra…
地址:http://codeforces.com/contest/782/problem/C 题目: C. Andryusha and Colored Balloons time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Andryusha goes through a park each day. The squares and…
A. Andryusha and Socks time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Andryusha is an orderly boy and likes to keep things in their place. Today he faced a problem to put his socks in the…
B. The Meeting Place Cannot Be Changed time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates…
本文版权归ljh2000和博客园共有,欢迎转载,但须保留此声明,并给出原文链接,谢谢合作. 本文作者:ljh2000 作者博客:http://www.cnblogs.com/ljh2000-jump/转载请注明出处,侵权必究,保留最终解释权! 题目链接:codeforces781A Andryusha and Colored Balloons 正解:构造+结论 解题报告: 考虑答案显然是$max(度数)+1$,这个似乎很好想,对于每个点我需要保证与之相邻的所有点颜色互不相同且与自己不同,那么需要…
C - Andryusha and Colored Balloons 思路: 水题: 代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define maxn 200005 ],V[maxn<<],dis[maxn],ans; void dfs(int now,int fa) { ; for…
C. Andryusha and Colored Balloons time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Andryusha goes through a park each day. The squares and paths between them look boring to Andryusha, so he…
试题 算法训练 Balloons in a Box 问题描述 你要写一个程序,使得能够模拟在长方体的盒子里放置球形的气球. 接下来是模拟的方案.假设你已知一个长方体的盒子和一个点集.每一个点代表一个可以放置气球的位置.在一个点上放置一个气球,就是以这个点为球心,然后让这个球膨胀,直到触及盒子的边缘或者一个之前已经被放置好的气球.你不能使用一个在盒子外面或者在一个之前已经放置好的气球里面的点.但是,你可以按你喜欢的任意顺序使用这些点,而且你不需要每个点都用.你的目标是按照某种顺序在盒子里放置气球,…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/burst-balloons/description/ 题目描述: Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are aske…
C. Andryusha and Colored Balloons time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Andryusha goes through a park each day. The squares and paths between them look boring to Andryusha, so he…
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图的遍历的定义: 从图的某个顶点出发访问遍图中所有顶点,且每个顶点仅被访问一次.(连通图与非连通图) 深度优先遍历(DFS): 1.访问指定的起始顶点: 2.若当前访问的顶点的邻接顶点有未被访问的,则任选一个访问之:反之,退回到最近访问过的顶点:直到与起始顶点相通的全部顶点都访问完毕: 3.若此时图中尚有顶点未被访问,则再选其中一个顶点作为起始顶点并访问之,转 2: 反之,遍历结束. 连通图的深度优先遍历类似于树的先根遍历 如何判别V的邻接点是否被访问? 解决办法:为每个顶点设立一个“访问标志”…
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[算法入门] 郭志伟@SYSU:raphealguo(at)qq.com 2012/05/12 1.前言 深度优先搜索(缩写DFS)有点类似广度优先搜索,也是对一个连通图进行遍历的算法.它的思想是从一个顶点V0开始,沿着一条路一直走到底,如果发现不能到达目标解,那就返回到上一个节点,然后从另一条路开始走到底,这种尽量往深处走的概念即是深度优先的概念. 你可以跳过第二节先看第三节,:) 2.深度优先搜索VS广度优先搜索 2.1演示深度优先搜索的过程 还是引用上篇文章的样例图,起点仍然是V0,我们修…
There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of s…
Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and r…