HDU-4292-Food(最大流,Dinic)

Posted 2022-06-25 ydddd

链接:

https://vjudge.net/problem/HDU-4292

题意:

　　You, a part-time dining service worker in your college’s dining hall, are now confused with a new problem: serve as many people as possible.
　　The issue comes up as people in your college are more and more difficult to serve with meal: They eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly.
　　You have prepared F (1 <= F <= 200) kinds of food and D (1 <= D <= 200) kinds of drink. Each kind of food or drink has certain amount, that is, how many people could this food or drink serve. Besides, You know there’re N (1 <= N <= 200) people and you too can tell people’s personal preference for food and drink.
　　Back to your goal: to serve as many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service.

思路:

最大流,建图,模板题.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;

const int MAXN = 200+10;
const int INF = 1e9;

struct Edge

    int from, to, cap;
;
vector<int> G[MAXN*4];
vector<Edge> edges;
int Dis[MAXN*4];
int Fo, Dr;
int n, f, d, s, t;

void Init()

    for (int i = s;i <= t;i++)
        G[i].clear();
    edges.clear();


void AddEdge(int from, int to, int cap)

    edges.push_back(Edgefrom, to, cap);
    edges.push_back(Edgeto, from, 0);
    G[from].push_back(edges.size()-2);
    G[to].push_back(edges.size()-1);


bool Bfs()

    memset(Dis, -1, sizeof(Dis));
    queue<int> que;
    que.push(s);
    Dis[s] = 0;
    while (!que.empty())
    
        int u = que.front();
        que.pop();
        for (int i = 0;i < G[u].size();i++)
        
            Edge &e = edges[G[u][i]];
            if (e.cap > 0 && Dis[e.to] == -1)
            
                Dis[e.to] = Dis[u]+1;
                que.push(e.to);
            
        
    
    return Dis[t] != -1;


int Dfs(int u, int flow)

    if (u == t)
        return flow;
    int res = 0;
    for (int i = 0;i < G[u].size();i++)
    
        Edge &e = edges[G[u][i]];
        if (e.cap > 0 && Dis[u]+1 == Dis[e.to])
        
            int tmp = Dfs(e.to, min(flow, e.cap));
            e.cap -= tmp;
            flow -= tmp;
            res += tmp;
            edges[G[u][i]^1].cap += tmp;
            if (flow == 0)
                break;
        
    
    if (res == 0)
        Dis[u] = -1;
    return res;


int MaxFlow()

    int res = 0;
    while (Bfs())
        res += Dfs(s, INF);
    return res;


int main()

    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> n >> f >> d)
    
        s = 0, t = n*2+f+d+1;
        Init();
        for (int i = 1;i <= f;i++)
        
            cin >> Fo;
            AddEdge(0, 2*n+i, Fo);
        
        for (int i = 1;i <= d;i++)
        
            cin >> Dr;
            AddEdge(2*n+f+i, t, Dr);
        
        for (int i = 1;i <= n;i++)
            AddEdge(i*2-1, i*2, 1);
        char ok;
        for (int i = 1;i <= n;i++)
        
            for (int j = 1;j <= f;j++)
            
                cin >> ok;
                if (ok == 'Y')
                    AddEdge(2*n+j, 2*i-1, 1);
            
        
        for (int i = 1;i <= n;i++)
        
            for (int j = 1;j <= d;j++)
            
                cin >> ok;
                if (ok == 'Y')
                    AddEdge(2*i, 2*n+f+j, 1);
            
        
        int res = MaxFlow();
        cout << res << endl;
    

    return 0;