D. Cow and Snacks
D. Cow and Snacks 传送门
The legendary Farmer John is throwing a huge party, and animals from all over the world are hanging out at his house. His guests are hungry, so he instructs his cow Bessie to bring out the snacks! Moo!
There are n
snacks flavors, numbered with integers 1,2,…,n. Bessie has n snacks, one snack of each flavor. Every guest has exactly two favorite flavors. The procedure for eating snacks will go as follows:
First, Bessie will line up the guests in some way.Then in this order, guests will approach the snacks one by one.Each guest in their turn will eat all remaining snacks of their favorite flavor. In case no favorite flavors are present when a guest goes up, they become very sad.Help Bessie to minimize the number of sad guests by lining the guests in an optimal way.
Input
The first line contains integers n and k (2≤n≤1e5, 1≤k≤1e5), the number of snacks and the number of guests.
The i-th of the following k lines contains two integers xi and yi (1≤xi,yi≤n, xi≠yi), favorite snack flavors of the i-th guest.
Output
Output one integer, the smallest possible number of sad guests.
Examples
Input
5 4 1 2 4 3 1 4 3 4
Output
1
Input
6 5 2 3 2 1 3 4 6 5 4 5
Output
0
Note
In the first example, Bessie can order the guests like this: 3,1,2,4. Guest 3 goes first and eats snacks 1 and 4. Then the guest 1 goes and eats the snack 2 only, because the snack 1 has already been eaten. Similarly, the guest 2 goes up and eats the snack 3 only. All the snacks are gone, so the guest 4 will be sad. In the second example, one optimal ordering is 2,1,3,5,4. All the guests will be satisfied.
思路： 根据每个客人爱吃的零食的种类编号，构建一条边。每条边的端点是每种口味零食的编号。我们把图建好之后, 枚举每个客人爱吃的零食的口味编号， 以两种口味为起点进行dfs每遍历到一个端点（没有被标记过的）就会满足一个客人。
时间复杂度O(n+k).
AC代码：
#include <bits/stdc++.h>
using namespace std;
#define io ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define mset(a, n) memset(a, n, sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<int,PII> PPI;
typedef pair<ll, ll> PLL;
const int maxn = 1e5+5;
const ll Mod = 998244353;
int n,k;
bool vis[maxn];
int ans;
int u[maxn], v[maxn];
vector<vector<int> > G;
void dfs(int s){
int len = G[s].size();
for(int i=0;i<len;i++){
int to = G[s][i];
if(vis[to]) continue;
vis[to] = true; ans -= 1;
dfs(to);
}
}
int main(){
scanf("%d%d", &n, &k);
G.clear();
G.resize(n+1);
for(int i=1;i<=k;i++){
scanf("%d%d", &u[i], &v[i]);
G[u[i]].push_back(v[i]);
G[v[i]].push_back(u[i]);
}
ans = k;
mset(vis, false);
for(int i=1;i<=k;i++){
if(!vis[u[i]]){
vis[u[i]] = true; vis[v[i]] = true; ans -= 1;
dfs(u[i]);
dfs(v[i]);
}
}
printf("%dn", ans);
return 0;
}
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