HDU 3215 Being a Hero（最小割）

HDU 3215 Being a Hero

题目链接

题意：一个英雄，分到几个城市，每一个城市有一个价值，可是要求分到城市后，必须破坏掉道路使得首都1都不能到达，破坏道路有开销。问最大能获得的收益和须要破坏的道路ID

思路：最小割，城市1做源点。有向边建图，容量为代价，然后每一个能够分的城市连到汇点。容量为价值，跑一下最小割就可以

代码：

 

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int MAXNODE = 1005;
const int MAXEDGE = 300005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
	int u, v, id;
	Type cap, flow;
	Edge() {}
	Edge(int u, int v, Type cap, Type flow, int id) {
		this->u = u;
		this->v = v;
		this->cap = cap;
		this->flow = flow;
		this->id = id;
	}
};

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, int id) {
		edges[m] = Edge(u, v, cap, 0, id);
		next[m] = first[u];
		first[u] = m++;
		edges[m] = Edge(v, u, 0, 0, id);
		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;
	}

	Type 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);
		}
		return flow;
	}

	void MinCut() {
		cut.clear();
		for (int i = 0; i < m; i += 2) {
			if (vis[edges[i].u] && !vis[edges[i].v] && edges[i].id)
				cut.push_back(i);
		}
		int sz = cut.size();
		printf("%d", sz);
		for (int i = 0; i < sz; i++)
			printf(" %d", edges[cut[i]].id);
		printf("\n");
	}
} gao;

int t, n, m, f;

int main() {
	int cas = 0;
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d%d", &n, &m, &f);
		gao.init(n + 1);
		int u, v, w, tot = 0;
		for (int i = 1; i <= m; i++) {
			scanf("%d%d%d", &u, &v, &w);
			gao.add_Edge(u, v, w, i);
		}
		while (f--) {
			scanf("%d%d", &u, &w);
			tot += w;
			gao.add_Edge(u, 0, w, 0);
		}
		printf("Case %d: %d\n", ++cas, tot - gao.Maxflow(1, 0));
		gao.MinCut();
	}
	return 0;
}