洛谷P1892 [BOI2003]团伙

并查集模板题。对于敌对关系有两种解决方案，

一是另开一个数组存敌人，这个数组无需预处理但是处理关系时需要特判，小心与0的合并；

还有一种巧妙的办法是把fa[]数组开两倍空间，fa[i]的敌人设为fa[i+n]，需要预处理然而处理关系时直接合并就行了。

code：

 1 #include<iostream>
 2 #include<cstdio>
 3 
 4 using namespace std;
 5 
 6 const int Maxn = 1010;
 7 
 8 int fa[Maxn],ene[Maxn],vis[Maxn];
 9 int find(int x){
10     if(fa[x] == x)return x;
11     return fa[x] = find(fa[x]);
12 }
13 int n,m,p,q,ans;
14 char ch;
15 
16 int main(){
17     cin >> n >> m;
18     for(int i = 1;i <= n;i++)fa[i] = i;
19     for(int i = 1;i <= m;i++){
20         cin >> ch >> p >> q;
21         if(ch == 'E'){
22             if(ene[p])fa[find(q)] = find(ene[p]);
23             else ene[p] = find(q);
24             if(ene[q])fa[find(p)] = find(ene[q]);
25             else ene[q] = find(p);
26         }
27         else fa[find(p)] = find(q);
28     }
29     for(int i = 1;i <= n;i++)if(!vis[find(i)])vis[find(i)] = 1,ans++;//vis[find(i)]
30     cout << ans;
31 return 0;
32 }

 1 #include<iostream>
 2 #include<cstdio>
 3 
 4 using namespace std;
 5 
 6 const int Maxn = 1010;
 7 int fa[Maxn<<1];
 8 
 9 int find(int x){
10     if(fa[x] == x)return x;
11     return fa[x] = find(fa[x]);
12 }
13 
14 int n,x,y,m,ans;
15 char ch;
16 
17 int main(){
18     cin >> n >> m;
19     for(int i = 1;i <= n<<1;i++)fa[i] = i;
20     for(int i = 1;i <= m;i++){
21         cin >> ch >> x >> y;
22         if(ch == 'E'){
23             fa[find(x+n)] = find(y);
24             fa[find(y+n)] = find(x);
25         }
26         else fa[find(x)] = find(y);
27     }
28     for(int i = 1;i <= n;i++)if(fa[i] == i)ans++;
29     cout << ans;
30 return 0;
31 }