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POJ 1637 Sightseeing tour

题目链接

题意：给一些有向边一些无向边，问是否能把无向边定向之后确定一个欧拉回路

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

先把有向边任意定向了，然后根据每个点的入度出度之差，可以确定每个点需要调整的次数，然后中间就是需要调整的边，容量为1，这样去建图最后判断从源点出发的边是否都满流即可

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;
}

POJ 1637 Sightseeing tour（最大流）

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原文地址：http://blog.csdn.net/accelerator_/article/details/40918977