poj 3281 Dining

最新推荐文章于 2020-06-13 19:48:37 发布

转载最新推荐文章于 2020-06-13 19:48:37 发布 · 142 阅读

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原文链接：http://www.cnblogs.com/GadyPu/p/4773513.html

本文探讨了如何使用最大流算法解决一个有趣的实际问题：如何在考虑每头牛的食物和饮料偏好下，最大化满足牛群饮食需求的数量。通过构建一个特殊的网络流图，该文详细解释了算法实现过程，包括初始化网络、构建边和顶点，以及执行最大流搜索。

题目连接

http://poj.org/problem?id=3281

Dining

Description

Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others.

Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although he might not be able to stuff everybody, he wants to give a complete meal of both food and drink to as many cows as possible.

Farmer John has cooked F (1 ≤ F ≤ 100) types of foods and prepared D (1 ≤ D ≤ 100) types of drinks. Each of his N (1 ≤ N ≤ 100) cows has decided whether she is willing to eat a particular food or drink a particular drink. Farmer John must assign a food type and a drink type to each cow to maximize the number of cows who get both.

Each dish or drink can only be consumed by one cow (i.e., once food type 2 is assigned to a cow, no other cow can be assigned food type 2).

Input

Line 1: Three space-separated integers: N, F, and D
Lines 2..N+1: Each line i starts with a two integers F_i and D_i, the number of dishes that cow i likes and the number of drinks that cow i likes. The next F_i integers denote the dishes that cow i will eat, and the D_i integers following that denote the drinks that cow i will drink.

Output

Line 1: A single integer that is the maximum number of cows that can be fed both food and drink that conform to their wishes

Sample Input

4 3 3
2 2 1 2 3 1
2 2 2 3 1 2
2 2 1 3 1 2
2 1 1 3 3

Sample Output

最大流，拆点构图。

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
using std::min;
using std::sort;
using std::pair;
using std::swap;
using std::queue;
using std::vector;
using std::multimap;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 1; i <= (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int Max_N = 1100;
const int INF = 0x3f3f3f3f;
struct Ford_Fulkerson {
    struct edge { int to, cap, next, rev; }G[Max_N << 2];
    int  N, F, D, tot, head[Max_N];
    bool vis[Max_N], Food[Max_N][Max_N], Drink[Max_N][Max_N];
    inline void init(int n, int f, int d) {
        tot = 0;
        this->N = n, this->F = f, this->D =d;
        cls(head, -1), cls(Food, false), cls(Drink, false);
    }
    inline void add_edge(int u, int v, int cap = 1) {
        G[tot] = (edge){ v, cap, head[u], tot + 1 }; head[u] = tot++;
        G[tot] = (edge){ u,   0, head[v], tot - 1 }; head[v] = tot++;
    }
    inline void built() {
        int x, y, v;
        rep(i, N) {
            scanf("%d %d", &x, &y);
            while(x--) {
                scanf("%d", &v);
                Food[i][v] = true;
            }
            while(y--) {
                scanf("%d", &v);
                Drink[i][v] = true;
            }
        }
    }
    inline int dfs(int u, int t, int f) {
        if(u == t) return f;
        vis[u] = true;
        for(int i = head[u]; ~i; i = G[i].next) {
            edge &e = G[i];
            if(e.cap > 0 && !vis[e.to]) {
                int d = dfs(e.to, t, min(e.cap, f));
                if(d > 0) {
                    e.cap -= d;
                    G[e.rev].cap +=d;
                    return d;
                }
            }
        }
        return 0;
    }
    inline void max_flow(int s, int t) {
        int flow = 0;
        while(true) {
            cls(vis, false);
            int f = dfs(s, t, INF);
            if(!f) break;
            flow += f;
        }
        printf("%d\n", flow);
    }
    inline void solve() {
        int s = 1, t = s + F + 2 * N + D + 1;
        rep(i, F) {
            add_edge(s, s + i);
        }
        rep(i, D) {
            add_edge(s + F + 2 * N + i, t);
        }
        rep(i, N) {
            add_edge(s + F + i, s + F + N + i, 1);
            rep(j, F) {
                if(Food[i][j]) add_edge(s + j, s + F + i);
            }
            rep(j, D) {
                if(Drink[i][j]) add_edge(s + F + N + i, s + F + 2 * N + j);
            }
        }
        max_flow(s, t);
    }
}go;
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w+", stdout);
#endif
    int n, f, d;
    while(~scanf("%d %d %d", &n, &f, &d)) {
        go.init(n ,f ,d);
        go.built();
        go.solve();
    }
    return 0;
}

转载于:https://www.cnblogs.com/GadyPu/p/4773513.html