B - Dining （最大流）

最新推荐文章于 2020-07-11 23:19:46 发布

原创最新推荐文章于 2020-07-11 23:19:46 发布 · 358 阅读

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在一个农场中，每头奶牛都有特定的食物和饮料偏好。农民约翰准备了多种食物和饮料，但需要确保尽可能多的奶牛得到它们喜欢的完整餐食。此问题转化为一种匹配问题，通过构建图并使用最大流算法来解决，以找到最优的食物和饮料分配方案。

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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).

题意：匹配题，割点然后跑最大流。。。。。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#include <string>
#include <cmath>
#include <map>
#include <set>
using namespace std;
struct no{
    int i,j,v;
    int x[20],y[20];
};no a[1010],b[1010];
struct dd{
    int e,v,n,w;
};dd d[10010];
int top,h[1010],f[1010];
struct nn{
    int x,y,v;
};
void s0(int s,int e,int v){
    d[top].e=e;
    d[top].v=v;
    d[top].w=v;
    d[top].n=h[s];
    h[s]=top++;
}

void init(){
    top=0;
    memset(h,-1,sizeof(h));
}

int bfs(int s,int e){
    memset(f,-1,sizeof(f));
    queue<int>q;
    f[s]=0;
    q.push(s);
    while(!q.empty()){
        int p=q.front();
        q.pop();
        for(int i=h[p];i!=-1;i=d[i].n){
            int y=d[i].e;
            if(f[y]==-1&&d[i].v>0){
                f[y]=f[p]+1;
                q.push(y);
            }
        }
    }
    return f[e]!=-1;
}

int dfs(int k,int n,int v){
    if(k==n)return v;
    int v1=0,anq=0;
    for(int i=h[k];i!=-1;i=d[i].n){
        int y=d[i].e;
        if(f[y]!=f[k]+1||d[i].v<=0)continue;
        anq=dfs(y,n,min(v,d[i].v));
        if(!anq)continue;
        v-=anq;
        v1+=anq;
        d[i].v-=anq;
        d[i^1].v+=anq;
        if(!v)break;
    }
    return v1;
}

int main()
{
    int m,n1,n2;
    while(scanf("%d%d%d",&m,&n1,&n2)!=EOF){
        init();
        int k1,k2,e;
        for(int i=201;i<=200+n1;i++)
            s0(0,i,1),s0(i,0,0);
        for(int i=301;i<=300+n2;i++)
            s0(i,401,1),s0(401,i,0);
        for(int i=1;i<=m;i++){
            s0(i*2-1,i*2,1);
            s0(i*2,i*2-1,1);
            scanf("%d%d",&k1,&k2);
            for(int j=1;j<=k1;j++){
                scanf("%d",&e);
                s0(e+200,i*2-1,1),s0(i*2-1,e+200,0);
            }
            for(int j=1;j<=k2;j++){
                scanf("%d",&e);
                s0(i*2,e+300,1),s0(e+300,i*2,0);
            }
        }
        int ans=0;
        while(bfs(0,401))ans+=dfs(0,401,1e8);
        printf("%d\n",ans);
    }
    return 0;
}