题意：求两条路 能从 400.0 -> 789.0 且这两条路不想交（除了端点400，789 ）

求只能走一次的网络流需要用到拆点，

将点i  拆成 i 和 i+n  i->i+n的容量为经过的次数  （这题为1 ）

若i 能到达 j  则连接 i+n-> j

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <time.h>;
#define cler(arr, val)    memset(arr, val, sizeof(arr))
#define FOR(i,a,b)  for(int i=a;i<=b;i++)
#define IN   freopen ("in.txt" , "r" , stdin);
#define OUT  freopen ("out.txt" , "w" , stdout);
typedef long long  LL;
const int MAXN = 514;
const int MAXM = 40101;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
struct Edge
{int to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void init()
{tol = 0;memset(head,-1,sizeof (head));
}
void addedge (int u,int v,int w,int rw = 0)
{edge[tol].to = v;edge[tol].cap = w;edge[tol].flow = 0;edge[tol].next = head[u];head[u] = tol++;edge[tol].to = u;edge[tol].cap = rw;edge[tol].flow = 0;edge[tol].next = head[v];head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{memset(dep,-1,sizeof(dep));memset(gap,0,sizeof(gap));gap[0] = 1;int front = 0, rear = 0;dep[end] = 0;Q[rear++] = end;while(front != rear){int u = Q[front++];for(int i = head[u]; i !=  -1; i = edge[i].next){int v = edge[i]. to;if(dep[v] != -1)continue;Q[rear++] = v;dep[v] = dep[u] + 1;gap[dep[v]]++;}}
}
int S[MAXN];
int sap(int start,int end, int N)
{BFS(start,end);memcpy(cur,head,sizeof(head));int top = 0;int u = start;int ans = 0;int i;while(dep[start] < N){if(u == end){int Min = INF;int inser;for( i = 0; i < top; i++){if(Min > edge[S[i]].cap - edge[S[i]].flow){Min = edge[S[i]].cap - edge[S[i]].flow;inser = i;}}for( i = 0; i < top; i++){edge[S[i]]. flow += Min;edge[S[i]^1].flow -= Min;}ans += Min;top = inser;u = edge[S[top]^1].to;continue;}bool flag =  false;int v;for( i = cur[u]; i != -1; i = edge[i]. next){v = edge[i]. to;if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]){flag =  true;cur[u] = i;break;}}if(flag){S[top++] = cur[u];u = v;continue;}int Min = N;for( i = head[u]; i !=  -1; i = edge[i].next){if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min){Min = dep[edge[i].to];cur[u] = i;}}gap[dep[u]]--;if(!gap[dep[u]]) return ans;dep[u] = Min + 1;gap[dep[u]]++;if(u != start)u = edge[S[--top]^1].to;}return ans;
}
double gao(double x,double y,double a,double b)
{return sqrt((x-a)*(x-a)+(y-b)*(y-b));
}
struct point
{double f,x,y,r;point(){};
}a[202];
bool cmp(point a,point b)
{return a.f<b.f;
}
int main()
{
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);
#endifint t,n;scanf("%d",&t);while(t--){init();scanf("%d",&n);for(int i=0;i<n;i++)scanf("%lf%lf%lf%lf",&a[i].f,&a[i].x,&a[i].y,&a[i].r);sort(a,a+n,cmp);addedge(0,n,1);addedge(n-1,2*n-1,1);for(int i=0;i<n;i++){addedge(i,i+n,1);for(int j=i+1;j<n;j++){if(a[i].f<a[j].f&&gao(a[i].x,a[i].y,a[j].x,a[j].y)<(a[i].r+a[j].r))addedge(i+n,j,1);}}if(sap(0,2*n-1,2*n)==2)puts("Game is VALID");else puts("Game is NOT VALID");}return 0;
}