POJ 1637 Sightseeing tour

题目链接

题意：给一些有向边一些无向边，问能否把无向边定向之后确定一个欧拉回路

思路：这题的模型很的巧妙，转一个http://blog.csdn.net/pi9nc/article/details/12223693

先把有向边随意定向了，然后依据每一个点的入度出度之差，能够确定每一个点须要调整的次数，然后中间就是须要调整的边，容量为1，这样去建图最后推断从源点出发的边是否都满流就可以

代码：

#include <cstdio>

#include <cstring>

#include <queue>

#include <algorithm>

using namespace std;

const int MAXNODE = 205;

const int MAXEDGE = 10005;

typedef int Type;

const Type INF = 0x3f3f3f3f;

struct Edge {

	int u, v;

	Type cap, flow;

	Edge() {}

	Edge(int u, int v, Type cap, Type flow) {

		this->u = u;

		this->v = v;

		this->cap = cap;

		this->flow = flow;

	}

};

struct Dinic {

	int n, m, s, t;

	Edge edges[MAXEDGE];

	int first[MAXNODE];

	int next[MAXEDGE];

	bool vis[MAXNODE];

	Type d[MAXNODE];

	int cur[MAXNODE];

	vector<int> cut;

	void init(int n) {

		this->n = n;

		memset(first, -1, sizeof(first));

		m = 0;

	}

	void add_Edge(int u, int v, Type cap) {

		edges[m] = Edge(u, v, cap, 0);

		next[m] = first[u];

		first[u] = m++;

		edges[m] = Edge(v, u, 0, 0);

		next[m] = first[v];

		first[v] = m++;

	}

	bool bfs() {

		memset(vis, false, sizeof(vis));

		queue<int> Q;

		Q.push(s);

		d[s] = 0;

		vis[s] = true;

		while (!Q.empty()) {

			int u = Q.front(); Q.pop();

			for (int i = first[u]; i != -1; i = next[i]) {

				Edge& e = edges[i];

				if (!vis[e.v] && e.cap > e.flow) {

					vis[e.v] = true;

					d[e.v] = d[u] + 1;

					Q.push(e.v);

				}

			}

		}

		return vis[t];

	}

	Type dfs(int u, Type a) {

		if (u == t || a == 0) return a;

		Type flow = 0, f;

		for (int &i = cur[u]; i != -1; i = next[i]) {

			Edge& e = edges[i];

			if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {

				e.flow += f;

				edges[i^1].flow -= f;

				flow += f;

				a -= f;

				if (a == 0) break;

			}

		}

		return flow;

	}

	bool Maxflow(int s, int t) {

		this->s = s; this->t = t;

		Type flow = 0;

		while (bfs()) {

			for (int i = 0; i < n; i++)

				cur[i] = first[i];

			flow += dfs(s, INF);

		}

		for (int i = first[0]; i + 1; i = next[i])

			if (edges[i].flow != edges[i].cap) return false;

		return true;

	}

	void MinCut() {

		cut.clear();

		for (int i = 0; i < m; i += 2) {

			if (vis[edges[i].u] && !vis[edges[i].v])

				cut.push_back(i);

		}

	}

} gao;

const int N = 205;

const int M = 1005;

int t, n, m, in[N], out[N];

int u[M], v[M], w[M];

bool solve() {

	gao.init(n + 2);

	for (int i = 1; i <= n; i++) {

		if ((in[i] + out[i]) % 2) return false;

		if (in[i] > out[i]) gao.add_Edge(i, n + 1, (in[i] - out[i]) / 2);

		if (out[i] > in[i]) gao.add_Edge(0, i, (out[i] - in[i]) / 2);

	}

	for (int i = 0; i < m; i++) {

		if (w[i]) continue;

		gao.add_Edge(u[i], v[i], 1);

	}

	return gao.Maxflow(0, n + 1);

}

int main() {

	scanf("%d", &t);

	while (t--) {

		scanf("%d%d", &n, &m);

		memset(in, 0, sizeof(in));

		memset(out, 0, sizeof(out));

		for (int i = 0; i < m; i++) {

			scanf("%d%d%d", &u[i], &v[i], &w[i]);

			in[v[i]]++;

			out[u[i]]++;

		}

		printf("%s\n", solve() ? "possible" : "impossible");

	}

	return 0;

}