题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3018 Ant Trip Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1658    Accepted Submission(s): 641 Problem Description Ant Country consist of N to…

Ant Trip Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Ant Country consist of N towns.There are M roads connecting the towns. Ant Tony,together with his friends,wants to go through every part…

Ant Trip Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 904    Accepted Submission(s): 338 Problem Description Ant Country consist of N towns.There are M roads connecting the towns. Ant Tony,to…

Ant Trip Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3660    Accepted Submission(s): 1455 Problem Description Ant Country consist of N towns.There are M roads connecting the towns. Ant Tony,…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3018 题目大意:有n个点,m条边,人们希望走完所有的路,且每条道路只能走一遍.至少要将人们分成几组. 解题思路:先用并查集求出所有的连通块,然后判断每个连通块内点的度数,如果有奇数点则需要的组数ans+=奇数点/2:反之,所需组数ans+=1.注意:如果遇到孤立点即度数为0的点则不用算进去,因为没有跟他相连的边. 代码: #include<iostream> #include<cstdio&…

九野的博客,转载请注明出处:  http://blog.csdn.net/acmmmm/article/details/10858065 题意:n个点m条边的无向图,求用几笔可以把所有边画完(画过的边不再画) 思路: 并查集+欧拉回路 对于每个连通分量,若是欧拉回路则一笔画完,若不是则 需要: 奇度数点个数/2 然后把每个连通分量所需的笔数相加 这里要注意一个点是不用画的 #include<stdio.h> #include<algorithm> #include<iostr…

题目链接: Ant Trip Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 32768/32768 K (Java/Others) Problem Description   Ant Country consist of N towns.There are M roads connecting the towns. Ant Tony,together with his friends,wants to go through ev…

Ant Trip Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3524    Accepted Submission(s): 1393 Problem Description Ant Country consist of N towns.There are M roads connecting the towns. Ant Tony,…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3018 思路分析:题目可以看做一笔画问题,求最少画多少笔可以把所有的边画一次并且只画一次: 首先可以求出该无向图中连通图的个数,在每个无向连通图中求出需要画的笔数再相加即为所求.在一个无向连通图中,如果所有的点的度数为偶数则存在一个欧拉回路, 则只需要画一笔即可:如果图中存在度数为奇数的点,则需要画的笔数为度数为奇数的点的个数 /2:需要注意的孤立的点不需要画: 代码如下: #include <cst…

HDU - 3018 Ant Country consist of N towns.There are M roads connecting the towns. Ant Tony,together with his friends,wants to go through every part of the country. They intend to visit every road , and every road must be visited for exact one time.Ho…