题意：
T个测试数据
n个圆
下面 fre x y r 表示圆的频率 坐标和半径
要求：
从频率为400(最小的) 圆 走到频率为789(最大)的圆，再走回来，除起点每个点只能经过一次
问这样的路径是否存在
走法：从400->789时经过的圆频率只增不减， 只能走相交的圆
反之则频率只减不增，也只能走相交的圆
建图：
以789为源点， 400为汇点
其他点拆点拆成2个点，自己连自己，cap=1表示这个点只能走一次
然后跑一边最大流，当汇点流>=2时说明有这样的路径
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <set>
#include <vector>
#define N 605
#define inf 10000
#define ll int using namespace std;
inline ll Max(ll a,ll b){return a>b?a:b;}
inline ll Min(ll a,ll b){return a<b?a:b;}
struct node{
double fre;
int x,y, r;
}p[N], s, t;
bool Cross(node a,node b){
return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) < (a.r+b.r)*(a.r+b.r);
} struct Edge{
int from, to, cap, nex;
}edge[N*20];
int head[N], edgenum;
void addedge(int u, int v, int cap){
Edge E ={u, v, cap, head[u]};
edge[edgenum] = E;
head[u] = edgenum++;
} int sign[N*4];
bool BFS(int from, int to){
memset(sign, -1, sizeof(sign));
sign[from] = 0; queue<int>q;
q.push(from);
while( !q.empty() ){
int u = q.front(); q.pop();
for(int i = head[u]; i!=-1; i = edge[i].nex)
{
int v = edge[i].to;
if(sign[v]==-1 && edge[i].cap)
{
sign[v] = sign[u] + 1, q.push(v);
if(sign[to] != -1)return true;
}
}
}
return false;
}
int Stack[N*4], top, cur[N*4];
int dinic(int from, int to){ int ans = 0;
while( BFS(from, to) )
{
memcpy(cur, head, sizeof(head));
int u = from; top = 0;
while(1)
{
if(u == to)
{
int flow = inf, loc;//loc 表示 Stack 中 cap 最小的边
for(int i = 0; i < top; i++)
if(flow > edge[ Stack[i] ].cap)
{
flow = edge[Stack[i]].cap;
loc = i;
} for(int i = 0; i < top; i++)
{
edge[ Stack[i] ].cap -= flow;
edge[Stack[i]^1].cap += flow;
}
ans += flow;
top = loc;
u = edge[Stack[top]].from;
}
for(int i = cur[u]; i!=-1; cur[u] = i = edge[i].nex)//cur[u] 表示u所在能增广的边的下标
if(edge[i].cap && (sign[u] + 1 == sign[ edge[i].to ]))break;
if(cur[u] != -1)
{
Stack[top++] = cur[u];
u = edge[ cur[u] ].to;
}
else
{
if( top == 0 )break;
sign[u] = -1;
u = edge[ Stack[--top] ].from;
}
}
}
return ans;
}
ll n; int main(){
int T, i, j; scanf("%d",&T);
while(T--)
{
memset(head, -1, sizeof(head)); edgenum = 0; scanf("%d",&n);
for(i = 1; i <= n; i++)
{
scanf("%lf %d %d %d",&p[i].fre,&p[i].x,&p[i].y,&p[i].r);
if(p[i].fre == 400)
t = p[i],i--,n--;
else if(p[i].fre == 789)
s = p[i],i--,n--;
}
if(Cross(s,t)){printf("Game is VALID\n");continue;}
for(i = 1; i <= n; i++)
{
addedge(i,i+n,1);
addedge(i+n,i,0);
if(Cross(p[i],s))
{
addedge(0,i,1);
addedge(i,0,0);
}
if(Cross(p[i],t))
{
addedge(i+n, 2*n+10,1);
addedge(2*n+10,i+n,0);
}
for(j = 1; j <= n; j++)
if(Cross(p[i],p[j]) && i!=j)
{
if(p[i].fre>p[j].fre)
{
addedge(i+n,j,1);
addedge(j,i+n,0);
}
else
{
addedge(j+n,i,1);
addedge(i,j+n,0);
}
}
}
int ans = dinic(0, 2*n+10);
if(ans < 2)printf("Game is NOT VALID\n");
else printf("Game is VALID\n"); }
return 0;
}