POJ 1637 Sightseeing tour

最新推荐文章于 2019-04-22 23:33:00 发布

Wall_F

最新推荐文章于 2019-04-22 23:33:00 发布

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分类专栏： PKU

本文链接：https://blog.youkuaiyun.com/Wall_F/article/details/8237811

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本文介绍了一种利用Dinic算法解决混合图中欧拉回路问题的方法。通过构造特殊的图结构，并使用最大流算法来判断是否存在欧拉回路，为解决该类问题提供了一种新的思路。

大意不再赘述。

思路：

混合图的欧拉回路求解，具体步骤见另一篇博客：http://blog.youkuaiyun.com/wall_f/article/details/8237520

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;

const int MAXN = 1010;
const int MAXM = 50010;
const int INF = 0x3f3f3f3f;


struct Edge
{
	int v, f;
	int next;
}edge[MAXM];

int n, m;
int cnt;
int s, t;

int first[MAXN], level[MAXN];
int q[MAXN];
int ind[MAXN], outd[MAXN];
int totFlow;

void init()
{
	cnt = 0;
	totFlow = 0;
	memset(first, -1, sizeof(first));
	memset(ind, 0, sizeof(ind));
	memset(outd, 0, sizeof(outd));
}

void read(int u, int v, int f)
{
	edge[cnt].v = v, edge[cnt].f = f;
	edge[cnt].next = first[u], first[u] = cnt++;
}

void read_graph(int u, int v, int f)
{
	read(u, v, f);
	read(v, u, 0);
}

int bfs(int s, int t)
{
	memset(level, 0, sizeof(level));
	level[s] = 1;
	int front = 0, rear = 1;
	q[front] = s;
	while(front < rear)
	{
		int x = q[front++];
		if(x == t) return 1;
		for(int e  = first[x]; e != -1; e = edge[e].next)
		{
			int v = edge[e].v, f = edge[e].f;
			if(!level[v] && f)
			{
				level[v] = level[x] + 1;
				q[rear++] = v;
			}
		}
	}
	return 0;
}

int dfs(int u, int maxf, int t)
{
	if(u == t) return maxf;
	int ret = 0;
	for(int e = first[u]; e != -1; e = edge[e].next)
	{
		int v = edge[e].v, f = edge[e].f;
		if(level[v] == level[u] + 1 && f)
		{
			int Min = min(maxf-ret, f);
			f = dfs(v, Min, t);
			edge[e].f -= f;
			edge[e^1].f += f;
			ret += f;
			if(ret == maxf) return ret;
		}
	}
	return ret;
}

int Dinic(int s, int t)
{
	int ans = 0;
	while(bfs(s, t)) ans += dfs(s, INF, t);
	return ans;
}

void read_case()
{
	init();
	scanf("%d%d", &n, &m);
	while(m--)
	{
		int u, v, flag;
		scanf("%d%d%d", &u, &v, &flag);
		outd[u]++, ind[v]++;
		if(u != v)
		{
			if(!flag) read_graph(u, v, 1);
		}
	}
}

int build()
{
	int flag = 1;
	s = 0, t = n+1;
	for(int i = 1; i <= n; i++)
	{
		if((ind[i]+outd[i]) & 1) //出度加入度是奇数 
		{
			return 0;
		}
		else if(outd[i] > ind[i]) //出度大于入度 
		{
			int dif = outd[i]-ind[i];
			read_graph(s, i, dif/2);
			totFlow += dif/2;
			
		} //可能有入度等于出度的情况,连不连无所谓 
		else
		{
			int dif = ind[i]-outd[i];
			read_graph(i, t, dif/2);
		}
	}
	return 1;
}

void solve()
{
	read_case();
	int flag = build();
	int ans = Dinic(s, t);
	if(!flag) printf("impossible\n");
	else if(ans >= totFlow) printf("possible\n");
	else printf("impossible\n"); 
}

int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		solve();
	}
	return 0;
}