CodeForces 659E New Reform
题意:给你一个无向图,如今要求你把边改成有向的。 使得入度为0的点最少,输出有多少个点入度为0
思路:脑补一波结论。假设有环的话显然没有点入度为0,其余则至少有一个点入度为0,然后就DFS一波就能够了
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 100000+1000
#define LL long long
int cas=1,T;
vector<int> e[maxn];
int vis[maxn];
int flag=0;
void dfs(int u,int fa)
{
if (vis[u])
{
flag=1;
return;
}
vis[u]=1;
for (int i=0;i<e[u].size();i++)
{
int v = e[u][i];
if (v==fa)
continue;
dfs(v,u);
}
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for (int i = 1;i<=m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
e[u].push_back(v);
e[v].push_back(u);
}
int ans = 0;
for (int i = 1;i<=n;i++)
{
if (!vis[i])
{
flag = 0;
dfs(i,-1);
if (!flag)
ans++;
}
}
printf("%d\n",ans);
//freopen("in","r",stdin);
//scanf("%d",&T);
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
}
Description
Berland has n cities connected by m bidirectional roads. No road connects a city to itself, and each pair of cities is connected
by no more than one road. It is not guaranteed that you can get from any city to any other one, using only the existing roads.
The President of Berland decided to make changes to the road system and instructed the Ministry of Transport to make this reform. Now, each road should be unidirectional (only lead from one city to another).
In order not to cause great resentment among residents, the reform needs to be conducted so that there can be as few separate cities as possible. A city is considered separate, if no road leads into it, while it is
allowed to have roads leading from this city.
Help the Ministry of Transport to find the minimum possible number of separate cities after the reform.
Input
The first line of the input contains two positive integers, n and m — the number of the cities and the number of roads in Berland
(2 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000).
Next m lines contain the descriptions of the roads: the i-th road is determined by two distinct integers xi, yi (1 ≤ xi, yi ≤ n, xi ≠ yi),
where xi and yi are the numbers of the cities
connected by the i-th road.
It is guaranteed that there is no more than one road between each pair of cities, but it is not guaranteed that from any city you can get to any other one, using only roads.
Output
Print a single integer — the minimum number of separated cities after the reform.
Sample Input
4 3
2 1
1 3
4 3
1
5 5
2 1
1 3
2 3
2 5
4 3
0
6 5
1 2
2 3
4 5
4 6
5 6
1
Hint
In the first sample the following road orientation is allowed: ,
,
.
The second sample: ,
,
,
,
.
The third sample: ,
,
,
,
.
CodeForces 659E New Reform的更多相关文章
- Codeforces 659E New Reform【DFS】
题目链接: http://codeforces.com/problemset/problem/659/E 题意: 给定n个点和m条双向边,将双向边改为单向边,问无法到达的顶点最少有多少个? 分析: 无 ...
- CodeForces 659E New Reform (图的遍历判环)
Description Berland has n cities connected by m bidirectional roads. No road connects a city to itse ...
- [图中找环] Codeforces 659E New Reform
New Reform time limit per test 1 second memory limit per test 256 megabytes input standard input out ...
- codeforces 659E . New Reform 强连通
题目链接 对于每一个联通块, 如果有一个强连通分量, 那么这个联通块对答案的贡献就是0. 否则对答案贡献是1. #include <iostream> #include <vecto ...
- codeforces 659E E. New Reform(图论)
题目链接: E. New Reform time limit per test 1 second memory limit per test 256 megabytes input standard ...
- Codeforces 732F. Tourist Reform (Tarjan缩点)
题目链接:http://codeforces.com/problemset/problem/732/F 题意: 给出一个有n个点m条边的无向图,保证联通,现在要求将所有边给定一个方向使其变成有向图,设 ...
- CodeForces 732F Tourist Reform
边双连通分量. 这题有一点构造的味道.一个有向图,经过强连通缩点之后会形成一个有向无环图. 如果将最大的强连通分量放在顶端,其余的强连通分量都直接或间接指向他,那么这样就构造出了符合要求的图. 接下来 ...
- CodeForces 723E One-Way Reform
构造. 有一种十分巧妙的方法可以使图中所有度数为偶数的节点,经过每条边定向后,出度和入度都相等. 首先统计每个节点的度数,将度数为奇数的节点与编号为$n+1$的节点连边,这样一来,这张新图变成了每个节 ...
- Codeforces Round #375 (Div. 2) E. One-Way Reform 欧拉路径
E. One-Way Reform 题目连接: http://codeforces.com/contest/723/problem/E Description There are n cities a ...
随机推荐
- linux 安装 pip
# wget https://bootstrap.pypa.io/get-pip.py # python get-pip.py
- SQLite 字段数据类型
一般数据采用的固定的静态数据类型,而SQLite采用的是动态数据类型,会根据存入值自动判断. SQLite具有以下五种数据类型: 1.NULL:空值. 2.INTEGER:带符号的整型,具体取决有存入 ...
- windows 添加自助白名单
由于公司分部用的是动态IP,又需要用到总部的OA系统,OA完全开放对外不安全,所以写了这个工具 项目地址 https://github.com/cainiaoit/Windows-firewall-s ...
- 3.Centos-Docker-rancher
1.安装mysql,设置密码 docker run -d --name mysqldb -e MYSQL_ROOT_PASSWORD=密码 mysql:latest --character-set-s ...
- 【linux高级程序设计】(第十四章)TCP高级应用
文件I/O方式比较 1.阻塞式文件I/O 进程从调用函数开始,直到返回这段时间都处于阻塞状态. 2.非阻塞式文件I/O 如果当前没有数据可操作,将不阻塞当前进程,而是立即返回一个错误信息.需要反复尝试 ...
- 第一部分:MongoDB备忘录
一.NoSQL 简介 Nosql的全称是Not Only Sql,这个概念早起就有人提出,在09年的时候比较火.Nosql指的是非关系型数据库,而我们常用的都是关系型数据库.就像我们常用的mysql, ...
- ajax和json数据
一.Ajax概述 1.什么是同步,什么是异步 同步现象:客户端发送请求到服务器端,当服务器返回响应之前,客户端都处于等待 卡死状态 异步现象:客户端发送请求到服务器端,无论服务器是否返回响应, ...
- AC日记——小M的作物 bzoj 3438
3438 思路: 最小割(完全不懂看的题解): s向每个作物连边,s-x ai,x-t bi: 然后s向作物集合连边,cia: 作物集合拆点向t连边,cib: 作物集合第一个点向作物连边INF: 作物 ...
- 牛客网 牛客小白月赛5 I.区间 (interval)-线段树 or 差分数组?
牛客小白月赛5 I.区间 (interval) 休闲的时候写的,但是写的心情有点挫,都是完全版线段树,我的一个队友直接就水过去了,为啥我的就超内存呢??? 试了一晚上,找出来了,多初始化了add标记数 ...
- NEUQOJ 1999: 三角形or四边形?【搜索联通块/模拟】
http://newoj.acmclub.cn/problems/1999 1999: 三角形or四边形? 描述 题目描述: JiangYu很无聊,所以他拿钉子在板子上戳出了一个由.#组成的10*10 ...