HDU 4183 Pahom on Water 来回走不重复点的网络流

题目来源：HDU 4183 Pahom on Water

题意：若干个区域 每个区域有一个值 区域是圆 给出圆心和半径

从起点（值为400.0）到终点（值为789.0）满足走相交的圆 并且值必须递增 然后从终点到起点 值必须递减 此外区域只能去一次

思路：建图 相互能走的区域连一条边 因为只能走一次 所以拆点 如果没有来回 只有去 那么判断最大流为1即可

现在还要回来 并且回来的条件和去的条件想法（一个递增一个递减）可以反向考虑给源点cap=2 最大流为2

#include <cstdio>
#include <queue>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 610;
const int INF = 999999999;
struct Edge
{
	int from, to, cap, flow;
	Edge(){}
	Edge(int from, int to, int cap, int flow) : from(from), to(to), cap(cap), flow(flow){}
};

struct Point
{
	double d, x, y, r;
	Point(){}
	Point(double d, double x, double y, double r) : d(d), x(x), y(y), r(r){}
}a[maxn];
int n, m, s, t;
vector <Edge> edges;
vector <int> G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];void AddEdge(int from, int to, int cap)
{
	edges.push_back(Edge(from, to, cap, 0));
	edges.push_back(Edge(to, from, 0, 0));
	m = edges.size();
	G[from].push_back(m-2);
	G[to].push_back(m-1);
}
bool BFS()
{
	memset(vis, 0, sizeof(vis));
	queue <int> Q;
	Q.push(s);
	d[s] = 0;
	vis[s] = 1;
	while(!Q.empty())
	{
		int x = Q.front(); Q.pop();
		for(int i = 0; i < G[x].size(); i++)
		{
			Edge& e = edges[G[x][i]];
			if(!vis[e.to] && e.cap > e.flow)
			{
				vis[e.to] = 1;
				d[e.to] = d[x] + 1;
				Q.push(e.to);
			}
		}
	}
	return vis[t];
}
int DFS(int x, int a)
{
	if(x == t || a == 0)
		return a;
	int flow = 0, f;
	for(int& i = cur[x]; i < G[x].size(); i++)
	{
		Edge& e = edges[G[x][i]];
		if(d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap-e.flow))) > 0)
		{
			e.flow += f;
			edges[G[x][i]^1].flow -= f;
			flow += f;
			a -= f;
			if(a == 0)
				break;
		}
	}
	return flow;
}

int Maxflow()
{
	int flow = 0;
	while(BFS())
	{
		memset(cur, 0, sizeof(cur));
		flow += DFS(s, INF);
	}
	return flow;
}

bool ok(int i, int j)
{
	if(a[i].d >= a[j].d)
		return false;
	if((a[i].r + a[j].r)*(a[i].r + a[j].r) > (a[i].x - a[j].x)*(a[i].x - a[j].x)+(a[i].y - a[j].y)*(a[i].y - a[j].y))
		return true;
	return false;
}
int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		int k;
		scanf("%d", &k);
		n = k*2;
		edges.clear();
		for(int i = 0; i < n; i++)
			G[i].clear();
		for(int i = 0; i < k; i++)
		{
			scanf("%lf %lf %lf %lf", &a[i].d, &a[i].x, &a[i].y, &a[i].r);
			if(a[i].d == 400.0)
				s = i;
			if(a[i].d == 789.0)
				t = i;
		}
		for(int i = 0; i < k; i++)
		{
			if(i != s && i != t)
				AddEdge(i<<1, i<<1|1, 1);
			for(int j = 0; j < i; j++)
			{
				if(ok(i, j))
				{
					AddEdge(i<<1|1, j<<1, 1);
				}
				else if(ok(j, i))
				{
					AddEdge(j<<1|1, i<<1, 1);
				}
			}
		}
		AddEdge(s<<1, s<<1|1, 2);
		s = s<<1;
		t = t<<1;
		//printf("%d %d\n", s, t);
		if(Maxflow() >= 2)
			puts("Game is VALID");
		else
			puts("Game is NOT VALID");
	}
	return 0;
}

HDU 4183 Pahom on Water 来回走不重复点的网络流