Mouse Hunt

思路：利用并查集将会出现在同一个房间的下标统一化

在深度遍历中寻找同一房间的下标们消耗最小值

若是在老鼠跳转中出现a[i]=i，则连同同一房间的老鼠们都会去i房间

若是出现回环，则比较回环中的最小值

typedef long long ll;
typedef pair<int, int> p;
typedef unsigned long long ull;
int n;
int c[200005];
int a[200005];
int v[200005];
int visited[200005];
int fa[200005], rankv[200005];
int fin(int v) {
    return fa[v] = fa[v] == v ? v : fin(fa[v]);
}

ll ans = 0;
int dfs(int ind, int minc = 200000) { //遍历找到最后的顶点
    v[ind]++;
    // a[i]=i时或者出现a数组出现环时循环找出最小值
    if (v[ind] >= 3) {
    //    cout << v[ind] << endl;
        ans += minc;
        return 1;
    }
    if (v[ind] == 2) {
   //     cout << v[ind] << endl;
        dfs(a[ind], min(minc, c[ind]));
    }
    else {
   //     cout << v[ind] << endl;
        dfs(a[ind], minc);
    }
    return 0;
}

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        fa[i] = i;
        rankv[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
        cin >> c[i];
    }
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        int x = fin(i), y = fin(a[i]);
        if (rankv[x] < rankv[y]) {
            fa[x] = y;
        }
        else {
            fa[y] = x;
            if (rankv[y] == rankv[x]) {
                ++rankv[x];
            
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        if (visited[fin(i)] == 0) {
            visited[fin(i)] = 1;
            dfs(fin(i));
        }
    }
    printf("%lld
", ans);
    return 0;
}