POJ 3281 Dining(最大流）

来源：互联网发布：字体模仿软件编辑：程序博客网时间：2024/04/29 08:05

POJ 3281 Dining

题目链接

题意：n个牛，每个牛有一些喜欢的食物和饮料，每种食物饮料只有一个，问最大能匹配上多少只牛每个牛都能吃上喜欢的食物和喜欢的饮料

思路：最大流，建模源点到每个食物连一条边，容量为1，每个饮料向汇点连一条边容量为1，然后由于每个牛有容量1，所以把牛进行拆点，然后食物连向牛的入点，牛的出点连向食物，跑一下最大流即可

代码：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int MAXNODE = 505;const int MAXEDGE = 100005;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;}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])cut.push_back(i);}}} gao;int f, d, n;int main() {while (~scanf("%d%d%d", &n, &f, &d)) {int s = 0, t = f + d + n * 2 + 1;gao.init(t + 1);for (int i = 1; i <= f; i++)gao.add_Edge(s, i, 1);for (int i = 1; i <= d; i++)gao.add_Edge(f + i, t, 1);for (int i = 1; i <= n; i++) {gao.add_Edge(i + f + d, n + f + d + i, 1);int fn, dn;scanf("%d%d", &fn, &dn);int v;while (fn--) {scanf("%d", &v);gao.add_Edge(v, i + f + d, 1);}while (dn--) {scanf("%d", &v);gao.add_Edge(n + f + d + i, f + v, 1);}}printf("%d\n", gao.Maxflow(s, t));}return 0;}

1 0