hdu 3018 Ant Trip 【图论-欧拉路】
来源:互联网 发布:曼森家族 知乎 编辑:程序博客网 时间:2024/06/06 04:05
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 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 3
1 2
2 3
1 3
4 2
1 2
3 4
Sample Output
1
2
Hint
New ~~~ 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.
题目大意:在蚂蚁王国有n个城市,城市之间有m条路,托尼和他的朋友想游览每个城市,并且每个城市只去一次,让你算出,想要逛遍所有城市,最少需要几人?
AC代码:
# include <iostream># include <cstring># include <cstdio>using namespace std;# define MAXN 100005int father[MAXN];int degree[MAXN];int nodenum[MAXN];int odd[MAXN];int FindRt(int x){ if (-1 == father[x]) { return x; } else { return father[x] = FindRt(father[x]); }}void Unite(int a, int b){ int ta = FindRt(a); int tb = FindRt(b); if (ta != tb) { father[ta] = tb; }}int main(void){ int n, m; while (~scanf("%d %d", &n, &m)) { int i; memset(father, -1, sizeof(father)); memset(degree, 0, sizeof(degree)); memset(nodenum, 0, sizeof(nodenum)); memset(odd, 0, sizeof(odd)); for (i = 0; i < m; i++) { int u, v; scanf("%d %d", &u, &v); degree[u] ++; degree[v] ++; Unite(u, v); } for (i = 1; i <= n; i++) { nodenum[FindRt(i)] ++; //计算一个连通分量中的节点的个数 if (1 == degree[i] % 2) //如果为奇数度 { odd[FindRt(i)]++; //父节点的奇数度加一 } } int res = 0; for (i = 1; i <= n; i++) { if (nodenum[i] <= 1) { continue; } if (odd[i] == 0) //如果不是奇数度 则只需一人 { res++; } else if (odd[i] > 0) //如果是奇数度,则为奇数度/2 人 { res += (odd[i] / 2); } } printf("%d\n", res); } return 0;}
- hdu 3018 Ant Trip 【图论-欧拉路】
- hdu 3018 Ant Trip
- HDU-3018-Ant Trip
- hdu 3018 Ant Trip
- hdu 3018 Ant Trip
- hdu 3018 Ant Trip
- HDU 3018 Ant Trip
- HDU 3018 Ant Trip
- Hdu 3018 Ant Trip
- HDU 3018 Ant Trip
- HDU 3018 Ant Trip
- hdu 3018 Ant Trip
- HDU 3018Ant Trip
- hdu 3018 Ant Trip
- HDU 3018 Ant Trip HDU
- hdu acm 3018 Ant Trip
- hdu acm 3018 Ant Trip
- HDU 3018 Ant Trip(连通分量 & 欧拉路)
- Partially Observable Markov Decision Process部分可观察的马尔可夫决策过程
- AngularJS控制器
- java中JDBCf访问数据库时报java.sql.SQLException: ORA-01017: invalid username/password; logon denied的错误
- (13.1.2)PMBOK之一(附):组织结构文化及其影响,过程资产环境因素与项目、项目管理、产品生命周期
- Qt中用CSS对进度条的样式表进行设置
- hdu 3018 Ant Trip 【图论-欧拉路】
- 爬虫实习小结
- MultiDex精补篇,进一步知道MultiDex的配置
- Hive DDL
- File.io读取文件(七:总结)
- Python office编程:word和PowerPoint
- cookie与session简述
- hadoop开启后用http访问出错
- ZJOI2017二试成功爆炸记