【网络流】 HDU 4183 Pahom on Water 拆点

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_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    int 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;
}