题目大意：有N个点。每一个点都有对应的三维坐标(x,y,z)
如今要求每一个点都能获得水，或者水的方式有两种
1.自己挖井，费用为X * 海拔高度z
2.铺设管道引水。
a.假设海拔高度小于引水处。费用为两地曼哈顿距离*Y
b.假设海拔高度大于饮水处。费用为两地曼哈顿距离*Y + Z

解题思路：设置一个虚根。虚根引向全部的点，权值为挖井的费用，接着依照要求连边，求出最小树形图就可以

#include <cstdio>
#include <cstring>

#define N 1010
#define abs(a) ((a)>0?(a):(-(a)))

struct Edge{
    int u, v, c;
}E[N*N];

struct Point{
    int x, y, z;
}P[N];

int n, x, y, z, tot;

void AddEdge(int u, int v, int c) {
    E[tot].u = u; E[tot].v = v; E[tot++].c = c;
}

void init() {
    tot = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d%d%d", &P[i].x, &P[i].y, &P[i].z);
        AddEdge(0, i, P[i].z * x);
    }

    int k, t;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &k);
        while (k--) {
            scanf("%d", &t);
            if (t == i)
                continue;
            if (P[i].z >= P[t].z) {
                int dis = abs(P[t].x - P[i].x) + abs(P[t].y - P[i].y) + abs(P[t].z - P[i].z);
                AddEdge(i, t, dis * y);
            }
            else {
                int dis = abs(P[t].x - P[i].x) + abs(P[t].y - P[i].y) + abs(P[t].z - P[i].z);
                AddEdge(i, t, dis * y + z);
            }
        }
    }
}

#define INF 0x3f3f3f3f
int in[N], pre[N], vis[N], id[N];
int Directed_MST(int root, int n) {
    int ans = 0, u, v, tmp;
    while (1) {
        for (int i = 0; i < n; i++)
            in[i] = INF;
        for (int i = 0; i < tot; i++) {
            u = E[i].u;
            v = E[i].v;
            if (u != v && E[i].c < in[v]) {
                in[v] = E[i].c;
                pre[v] = u;
            }
        }
        memset(vis, -1, sizeof(vis));
        memset(id, -1, sizeof(id));
        in[root] = 0;
        int subnode = 0;
        for (int i = 0; i < n; i++) {
            ans += in[i];
            tmp = i;
            while (vis[tmp] != i && tmp != root && id[tmp] == -1) {
                vis[tmp] = i;
                tmp = pre[tmp];
            }
            if (id[tmp] == -1 && tmp != root) {
                u = pre[tmp];
                while (u != tmp) {
                    id[u] = subnode;
                    u = pre[u];
                }
                id[tmp] = subnode++;
            }
        }

        if (subnode == 0)
            break;
        for (int i = 0; i < n; i++)
            if (!(~id[i]))
                id[i] = subnode++;
        for (int i = 0; i < tot; i++) {
            tmp = E[i].v;
            E[i].u = id[E[i].u];
            E[i].v = id[E[i].v];

            if (E[i].u != E[i].v)
                E[i].c -= in[tmp];
        }
        n = subnode;
        root = id[root];
    }
    return ans;
}

void solve() {
    n++;
    int ans = Directed_MST(0,n);
    if (ans == -1) 
        printf("poor XiaoA\n");
    else
        printf("%d\n", ans);
}

int main() {
    while (scanf("%d%d%d%d", &n, &x, &y, &z) != EOF && n + x + y + z) {
        init();
        solve();
    }
    return 0;
}