HDOJ3018 Ant Trip(欧拉回路 + 并查集)
来源:互联网 发布:手机ar软件大全 编辑:程序博客网 时间:2024/05/16 08:13
Ant Trip
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2055 Accepted Submission(s): 796
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 of the country.
They intend to visit every road , and every road must be visited for exact one time.However,it may be a mission impossible for only one group of people.So they are trying to divide all the people into several groups,and each may start at different town.Now tony wants to know what is the least groups of ants that needs to form to achieve their goal.
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.However,it may be a mission impossible for only one group of people.So they are trying to divide all the people into several groups,and each may start at different town.Now tony wants to know what is the least groups of ants that needs to form to achieve their goal.
Input
Input contains multiple cases.Test cases are separated by several blank lines. Each test case starts with two integer N(1<=N<=100000),M(0<=M<=200000),indicating that there are N towns and M roads in Ant Country.Followed by M lines,each line contains two integers a,b,(1<=a,b<=N) indicating that there is a road connecting town a and town b.No two roads will be the same,and there is no road connecting the same town.
Output
For each test case ,output the least groups that needs to form to achieve their goal.
Sample Input
3 31 22 31 34 21 23 4
Sample Output
12HintNew ~~~ Notice: if there are no road connecting one town ,tony may forget about the town.In sample 1,tony and his friends just form one group,they can start at either town 1,2,or 3.In sample 2,tony and his friends must form two group.
经过每条边一次,每次一笔画,问你最少画多少次,孤立点不算其中。通过并查集判断连通集的数量,点的度数奇偶性判断。
笔画数 = 奇度数 + 欧拉回路数。
AC代码:
#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"#include "vector"using namespace std;const int MAXN = 1e5 + 5;int n, m, par[MAXN], deg[MAXN], odd[MAXN];bool vis[MAXN];vector<int> v;int find(int x){if(x == par[x]) return x;return par[x] = find(par[x]);}int main(int argc, char const *argv[]){while(scanf("%d%d", &n, &m) != EOF) {v.clear();memset(deg, 0, sizeof(deg));memset(odd, 0, sizeof(odd));memset(vis, false, sizeof(vis));for(int i = 1; i <= n; ++i) par[i] = i;for(int i = 0; i < m; ++i) {int a, b;scanf("%d%d", &a, &b);int fa = find(a), fb = find(b);deg[a]++, deg[b]++;par[fb] = fa;}for(int i = 1; i <= n; ++i) {int x = find(i);if(!vis[x]) {v.push_back(x);vis[x] = true;}if(deg[i] & 1) odd[x]++;}int ans = 0;for(int i = 0; i < v.size(); ++i) {int x = v[i];if(deg[x] == 0) continue;if(odd[x] == 0) ans++;else ans += odd[x] >> 1;}printf("%d\n", ans);}return 0;}
1 0
- HDOJ3018 Ant Trip(欧拉回路 + 并查集)
- hdoj3018 Ant Trip 欧拉回路
- HDU 3018 Ant Trip ( 并查集+欧拉回路 )
- hdu3018 Ant Trip--欧拉回路+并查集
- hdu-3018-Ant Trip(并查集&&欧拉回路)
- HDOJ3018 欧拉回路
- HDU 3018-Ant Trip(并查集&&欧拉)
- hdu3018 Ant Trip 欧拉回路
- HDU 3018 Ant Trip(欧拉回路)
- [欧拉回路] hdu 3018 Ant Trip
- [欧拉回路] hdu 3018 Ant Trip
- 【欧拉回路】 HDOJ 3018 Ant Trip
- hdu3018 Ant Trip 欧拉回路
- hdu 3018 Ant Trip 欧拉回路
- hdu3018 Ant Trip(欧拉回路)
- HDU3018 Ant Trip【欧拉回路】
- UVa302 - John's trip(并查集、欧拉回路、DFS)
- POJ1041 John's trip(欧拉回路 + 并查集 + dfs)
- 5道经典的程序题 (1)
- Processing - 练习(3)"飞碟追踪" - (渐进、延迟效果)
- 数据库三大范式
- LeetCode Merge Intervals
- HDU 5372 Segment Game 树状数组
- HDOJ3018 Ant Trip(欧拉回路 + 并查集)
- 周鸿祎:给那些仍旧在公司混日子的人
- javascript广告漂浮效果代码
- C++基础---函数的参数传递
- 3Sum
- 华为机试之数字倒序输出
- iOS开发之UI基础--KVC
- 小游戏
- 14.1 YAWL平台下项目部署和配置