源汇不确定时还原现场

有时候最大流问题需要我们枚举源点和汇点
此时每次枚举完毕的时候，要还原现场

• 对于边 $e \in E[\cdots \text{idx}]$，$e[i]$ 的流量 $|f|$ 为 $e[i \oplus 1]$ 的容量 $|c|$

• $\text{dinic}$ 算法维护的是残量网络 $G_f$，最开始流量为 $0$

所以还原现场 $2$ 步即可，注意残量网络中 $e = c(u, v)$，边维护的是容量
对于所有的边 $\textbf{for} \ \forall e_i \in E[\cdots \text{idx}]$

• 反向边容量退流，$e[i] \leftarrow e[i] + e[i \oplus 1]$

• 正向边流量清空，正向边流量等于反向边容量，所以 $e[i \oplus 1] \leftarrow 0$

• 对于冰块 $u$，它的每一条出边 $i$，必须满足 $\sum e_i \leqslant m_u$
  很自然地想到拆点，$u \to (u_1, u_2), \quad e(u_1, u_2) = m_u$

• 冰块 $(u, v)$ 之间能够互相跳到，充要条件是 $\text{dist}(u, v) \leqslant D$
  由于 $(u, v)$ 这条路径上可以允许无数企鹅经过
  （注意，对起跳终点 $v$ 并没有限制，对起点的限制已经加在拆点上了）
  添加边 $c(u, v) = \infty, \ c(v, u) = \infty, \quad u, v$ 是拆点后的

• 考虑源点，对于所有有企鹅的冰块，$\textbf{for} \ \forall i, \quad n_i > 0$:
  $c(S, i) = n_i$

• 枚举每一个汇点 $T \in [1 \cdots n]$，枚举前记得还原现场，如果
  $\textbf{dinic}(S, T) = tot, \quad res ++$

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
#include <cmath>
#include <bitset>
#include <assert.h>
#include <unordered_map>

using namespace std;
typedef long long ll;

#define Cmp(a, b) memcmp(a, b, sizeof(b))
#define Cpy(a, b) memcpy(a, b, sizeof(b))
#define Set(a, v) memset(a, v, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define _forS(i, l, r) for(set<int>::iterator i = (l); i != (r); i++)
#define _rep(i, l, r) for(int i = (l); i <= (r); i++)
#define _for(i, l, r) for(int i = (l); i < (r); i++)
#define _forDown(i, l, r) for(int i = (l); i >= r; i--)
#define debug_(ch, i) printf(#ch"[%d]: %d\n", i, ch[i])
#define debug_m(mp, p) printf(#mp"[%d]: %d\n", p->first, p->second)
#define debugS(str) cout << "dbg: " << str << endl;
#define debugArr(arr, x, y) _for(i, 0, x) { _for(j, 0, y) printf("%c", arr[i][j]); printf("\n"); }
#define _forPlus(i, l, d, r) for(int i = (l); i + d < (r); i++)
#define lowbit(i) (i & (-i))
#define MPR(a, b) make_pair(a, b)

pair<int, int> crack(int n) {
    int st = sqrt(n);
    int fac = n / st;
    while (n % st) {
        st += 1;
        fac = n / st;
    }

    return make_pair(st, fac);
}

inline ll qpow(ll a, int n) {
    ll ans = 1;
    for(; n; n >>= 1) {
        if(n & 1) ans *= 1ll * a;
        a *= a;
    }
    return ans;
}

template <class T>
inline bool chmax(T& a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a % b);
}

ll ksc(ll a, ll b, ll mod) {
    ll ans = 0;
    for(; b; b >>= 1) {
        if (b & 1) ans = (ans + a) % mod;
        a = (a * 2) % mod;
    }
    return ans;
}

ll ksm(ll a, ll b, ll mod) {
    ll ans = 1 % mod;
    a %= mod;

    for(; b; b >>= 1) {
        if (b & 1) ans = ksc(ans, a, mod);
        a = ksc(a, a, mod);
    }

    return ans;
}

template <class T>
inline bool chmin(T& a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool lexSmaller(vector<T> a, vector<T> b) {
    int n = a.size(), m = b.size();
    int i;
    for(i = 0; i < n && i < m; i++) {
        if (a[i] < b[i]) return true;
        else if (b[i] < a[i]) return false;
    }
    return (i == n && i < m);
}

// ============================================================== //

typedef pair<int, int> PII;
const int maxn = 210, maxm = (210 + 210 + 210*210) * 2;
const int inf = 0x3f3f3f3f;
const double eps = 1e-5;

int head[maxn], ver[maxm], e[maxm], ne[maxm], idx = 1;
int n, S = 0, T, tot = 0;
double D;
PII ice[maxn];

void add(int a, int b, int c) {
    ver[++idx] = b; e[idx] = c; ne[idx] = head[a]; head[a] = idx;
    ver[++idx] = a; e[idx] = 0; ne[idx] = head[b]; head[b] = idx;
}

bool check(const PII &a, const PII &b) {
    double dx = a.first - b.first, dy = a.second - b.second;
    return dx*dx + dy*dy <= D*D + eps;
}

void init() {
    memset(head, 0, sizeof head);
    idx = 1;
    tot = 0;
}

int d[maxn], cur[maxn];
bool bfs() {
    memset(d, -1, sizeof d);
    d[S] = 0, cur[S] = head[S];
    queue<int> q; q.push(S);
    while (q.size()) {
        int x = q.front(); q.pop();
        for (int i = head[x]; i; i = ne[i]) {
            int y = ver[i];
            if (d[y] == -1 && e[i]) {
                d[y] = d[x] + 1;
                cur[y] = head[y];
                if (y == T) return true;
                q.push(y);
            }
        }
    }
    return false;
}

int dinic(int u, int lim) {
    if (u == T) return lim;
    int flow = 0;
    for (int i = cur[u]; i && lim > flow; i = ne[i]) {
        cur[u] = i;
        int v = ver[i];
        if (d[v] == d[u] + 1 && e[i]) {
            int t = dinic(v, min(e[i], lim - flow));
            if (!t) d[v] = -1;
            flow += t, e[i] -= t, e[i^1] += t;
        }
    }
    return flow;
}

int dinic() {
    int res = 0, flow;
    while (bfs()) while (flow = dinic(S, inf)) res += flow;
    return res;
}

void solve() {
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        T = i;
        for (int k = 2; k <= idx; k += 2) {
            e[k] += e[k^1];
            e[k^1] = 0;
        }
        int res = dinic();
        if (res == tot) {
            cnt++;
            printf("%d ", i-1);
        }
    }
    if (cnt == 0) printf("-1\n");
    else printf("\n");
}

int main() {
    freopen("input.txt", "r", stdin);
    int kase;
    cin >> kase;
    while (kase--) {
        init();
        scanf("%d%lf", &n, &D);
        S = 0;
        // ice
        for (int i = 1; i <= n; i++) {
            int x, y, _n, _m;
            scanf("%d%d%d%d", &x, &y, &_n, &_m);
            ice[i] = {x, y};
            add(i, n+i, _m);
            add(S, i, _n);
            tot += _n;
        }
        // link node
        for (int i = 1; i <= n; i++) for (int j = i+1; j <= n; j++) {
            if (check(ice[i], ice[j])) {
                add(n+i, j, inf);
                add(n+j, i, inf);
            }
        }

        // then solve
        solve();
    }
}