傳送門
題意:有1~n種食物,k個客人,每個客人有兩種喜歡的食物,這些客人按照一定順序排隊喫飯,每個人輪到的時候會喫掉自己喜歡的,如果一個都沒有就不滿意,問最好的排隊方法可以讓多少客人不滿意
思路:題解原話“Since every animal has exactly two favorite snacks, this hints that we should model the problem as a graph.”,這種題目果然還要靠經驗。。一般這種事物間有相互關係的,很多都要用圖的模型來表示。食物爲節點,建好圖之後dfs跑一下就好了。。。
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<cstdio>
#include<queue>
#include<vector>
using namespace std;
const int MAXN = 1e5+5;
int n, k;
bool vis[MAXN];
bool gt[MAXN];
int head[MAXN], tot;
struct Edge{
int to, next;
int id;
}edge[MAXN*2];
void addEdge(int u, int v, int id){
edge[++tot].to = v;
edge[tot].next = head[u];
head[u] = tot;
edge[tot].id = id;
}
void dfs(int node){
vis[node] = true;
for(int i = head[node]; i != -1; i = edge[i].next){
int v = edge[i].to;
int id = edge[i].id;
if(vis[v] == false){
gt[id] = true;
dfs(v);
}
}
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
memset(head, -1, sizeof(head));
cin >> n >> k;
for(int i = 1; i <= k; ++i)
{
int u, v;
cin >> u >> v;
addEdge(u, v, i);
addEdge(v, u, i);
}
int ans = 0;
for(int i = 1; i <= n; ++i)
if(vis[i] == false) dfs(i);
for(int i = 1; i <= k; ++i)
if(gt[i] == false) ans++;
cout << ans << endl;
return 0;
}