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：

#include<iostream>
#include<cstdio>

using namespace std;

const int Maxn = 1010;

int fa[Maxn],ene[Maxn],vis[Maxn];
int find(int x){
    if(fa[x] == x)return x;
    return fa[x] = find(fa[x]);
}
int n,m,p,q,ans;
char ch;

int main(){
    cin >> n >> m;
    for(int i = 1;i <= n;i++)fa[i] = i;
    for(int i = 1;i <= m;i++){
        cin >> ch >> p >> q;
        if(ch == 'E'){
            if(ene[p])fa[find(q)] = find(ene[p]);
            else ene[p] = find(q);
            if(ene[q])fa[find(p)] = find(ene[q]);
            else ene[q] = find(p);
        }
        else fa[find(p)] = find(q);
    }
    for(int i = 1;i <= n;i++)if(!vis[find(i)])vis[find(i)] = 1,ans++;//vis[find(i)]
    cout << ans;
return 0;
}

#include<iostream>
#include<cstdio>

using namespace std;

const int Maxn = 1010;
int fa[Maxn<<1];

int find(int x){
    if(fa[x] == x)return x;
    return fa[x] = find(fa[x]);
}

int n,x,y,m,ans;
char ch;

int main(){
    cin >> n >> m;
    for(int i = 1;i <= n<<1;i++)fa[i] = i;
    for(int i = 1;i <= m;i++){
        cin >> ch >> x >> y;
        if(ch == 'E'){
            fa[find(x+n)] = find(y);
            fa[find(y+n)] = find(x);
        }
        else fa[find(x)] = find(y);
    }
    for(int i = 1;i <= n;i++)if(fa[i] == i)ans++;
    cout << ans;
return 0;
}