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.
Output
Print a single integer — the minimum number of separated cities after the reform.
Hint
In the first sample the following road orientation is allowed: , , .
The second sample: , , , , .
The third sample: , , , , .