您的位置:首页 > 其它

POJ 3281 Dining(最大流)

2014-11-06 17:32 302 查看


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;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: