题意:给你一个图 判断最少用几笔把这个图画出来(画过的边不能重新画) 思路: 并查集+欧拉回路 仔细想一想. 在一个强连通分量中 所有度为奇数的点之和÷2就是要画的笔画数 Now question :如果这是个欧拉回路怎么办? +1就OK啦 答案就是奇度数的点÷2+欧拉回路的个数 // by SiriusRen #include <cstdio> #include <cstring> #define N 100500 using namespace std; int ans,n,m…

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…

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…

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…

题意: Ant Tony和他的朋友们想游览蚂蚁国各地. 给你蚂蚁国的N个点和M条边,现在问你至少要几笔才能所有边都画一遍.(一笔画的时候笔不离开纸) 保证这M条边都不同且不会存在同一点的自环边. 也就是蚂蚁分组遍历整个无向图,他们试图把所有的人分成几个小组,每个小组可以从不同的城镇开始. Tony想知道最少需要几组.  Input输入包含多组测试用例,由多个空行分隔. 每个测试用例的第一行是两个整数N(1<=N<=100000).M(0<=M<=200000),表明蚂蚁国有N个城镇…

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) 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…

欧拉回路的题: 主要利用的是并查集,为了节省时间,压缩了它的路径: 代码: #include<cstdio> #include<cstring> #define maxn 100009 using namespace std; int f[maxn],rank[maxn],du[maxn]; bool vis[maxn],mark[maxn]; void make_set(int x) { f[x]=x; rank[x]=; } int find(int x) { int p=x,…

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,…