UVA 11248 - Frequency Hopping

题目链接

题意：给定一个网络，如今须要从1到N运输流量C，问是否可能，假设可能输出可能，假设不可能，再问能否通过扩大一条边的容量使得可能，假设能够输出这些边（按u先排再按v排），假设不行输出不可能

思路：先做一遍网络流，然后每次在最小割上进行添加容量，须要两个优化，每次找流量找到>= c就能够了，然后每次改动容量，能够直接从之前做过的网络流继续做就可以

代码：

#include <cstdio>

#include <cstring>

#include <queue>

#include <algorithm>

using namespace std;

const int MAXNODE = 105 * 2;

const int MAXEDGE = 100005;

typedef int Type;

const Type INF = 0x3f3f3f3f;

int n, m, c;

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;

	}

};

bool cmp(Edge a, Edge b) {

	if (a.u != b.u) return a.u < b.u;

	return a.v < b.v;

}

struct Dinic {

	int n, m, s, t;

	Edge edges[MAXEDGE], etmp[MAXEDGE];

	int first[MAXNODE];

	int next[MAXEDGE];

	bool vis[MAXNODE];

	Type d[MAXNODE];

	int cur[MAXNODE];

	vector<int> cut;

	vector<Edge> ans;

	Type flow;

	void init(int n) {

		this->n = n;

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

		m = 0;

		flow = 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;

	}

	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);

		}

	}

	bool Maxflow(int s, int t) {

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

		while (bfs()) {

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

				cur[i] = first[i];

			flow += dfs(s, INF);

			if (flow >= c) return true;

		}

		return false;

	}

	void solve() {

		if (Maxflow(0, n - 1) || c == 0) printf("possible\n");

		else {

			MinCut();

			ans.clear();

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

				etmp[i] = edges[i];

			int tmpf = flow;

			for (int i = 0; i < cut.size(); i++) {

				edges[cut[i]].cap = edges[cut[i]].flow + c;

				if (Maxflow(0, n - 1)) ans.push_back(edges[cut[i]]);

				flow = tmpf;

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

					edges[i] = etmp[i];

			}

			if (ans.size() == 0) printf("not possible\n");

			else {

				sort(ans.begin(), ans.end(), cmp);

				printf("possible option:");

				for (int i = 0; i < ans.size(); i++)

					printf("(%d,%d)%c", ans[i].u + 1, ans[i].v + 1, i == ans.size() - 1 ? '\n' : ',');

			}

		}

	}

} gao;

int main() {

	int cas = 0;

	while (~scanf("%d%d%d", &n, &m, &c) && n) {

		gao.init(n);

		int u, v, cap;

		while (m--) {

			scanf("%d%d%d", &u, &v, &cap);

			u--; v--;

			gao.add_Edge(u, v, cap);

		}

		printf("Case %d: ", ++cas);

		gao.solve();

	}

	return 0;

}