题解 | Dining-牛客假日团队赛2E题

题目描述

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).

输入描述:

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

输出描述:

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

示例1

输入

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

输出

3

说明

One way to satisfy three cows is:
Cow 1: no meal
Cow 2: Food #2, Drink #2
Cow 3: Food #1, Drink #1
Cow 4: Food #3, Drink #3
The pigeon-hole principle tells us we can do no better since there are only three kinds of food or drink. Other test data sets are more challenging, of course.

解答

这是一道匹配问题，我们可以用网络流建模来解决。
先考虑建立食物—牛—饮料的图，即：
1）：源点S向每种食物连容量为1的有向边
2）：每种食物向对应的牛（喜欢吃该食物的牛）连容量为1的有向边
3）：每头牛向喜欢喝的饮料连容量为1的有向边
4）：每种饮料向汇点T连容量为1的有向边
这种建模方式，思考一下，每种食物或饮料的确是只能供一头牛享用，因为受到了S到食物和饮料到T的容量1的限制，使得每个饮料和食物都只被用一次，但是，仔细想来，并未满足每头牛只享用一种食物和一种饮料。因为某头牛可能喜欢吃多种食物和饮料，而那头牛之前可能之前已匹配过一种食物和一种饮料，但可能存在之后的某次增广过程，是从那头牛喜欢吃的另一种食物进入牛这个点，且能从另一种饮料到达T，从而完成一次增广。
因而，我们还需加一类边：
5）：将每头牛拆成入点和出点，入点到出点连一条容量为1的有向边
这样，就能满足“每种食物或饮料只能供一头牛享用，且每头牛只享用一种食物和一种饮料”这两个条件了。
【代码】：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define MAXNODE 402
#define MAXE 4610
#define IMAX 214748364
struct EDGE{int f,t,next,flow;}a[MAXE];
int N,F,D,tot=0,last[MAXNODE],ans=0,S,T;
int d[MAXNODE],Q[MAXE]; 
bool map[MAXNODE][MAXNODE];
void add(int from,int to,int flow)
{
	a[tot].t=to;
	a[tot].flow=flow;
	a[tot].next=last[from];
	last[from]=tot++;
	a[tot].t=from;
	a[tot].flow=0;
	a[tot].next=last[to];
	last[to]=tot++;	
}
bool build()
{
	int right=0;
	for(int i=S;i<=T;i++)
	      d[i]=IMAX;
	d[S]=0;
	Q[++right]=S;
	for(int left=1;left<=right;left++)
	{
		  int now=Q[left];
		  for(int i=last[now];i!=-1;i=a[i].next)
		  {
		  	    int to=a[i].t;
		  	    if(a[i].flow && d[now]+1<d[to])
		  	    {
		  	    	  d[to]=d[now]+1;
		  	    	  if(to==T)   return true;
		  	    	  Q[++right]=to;
				}
		  }
	}
	return false;
}
int DINIC(int now,int flow)
{
	if(!flow)   return 0;
	if(now==T)  return flow;
	int rem=flow,use;
	for(int i=last[now];i!=-1;i=a[i].next)
	{
		  int to=a[i].t;
		  if(d[to]==d[now]+1 && (use=DINIC(to,min(flow,a[i].flow))))
		  {
		  	     a[i].flow-=use;
		  	     a[i^1].flow+=use;
		  	     flow-=use;
		  }
	}
	if(rem==flow)   d[now]=-1;
	return rem-flow;
}
void MaxFlow()
{
	int use;
	while(build())
	{
		  while(use=DINIC(S,IMAX))
		        ans+=use;
	}
}
int main()
{
	//freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	memset(last,-1,sizeof(last));
	scanf("%d%d%d",&N,&F,&D);
	S=0;T=F+2*N+D+1;
	for(int i=1;i<=F;i++)
	      add(S,i,1);	
	for(int i=1;i<=N;i++)
	      add(F+i,F+N+i,1);
	for(int i=1;i<=D;i++)
	      add(F+2*N+i,T,1);
	for(int i=1;i<=N;i++)
	{
		  int f,d;
		  scanf("%d%d",&f,&d);
		  for(int j=1;j<=f;j++)
		  {
		  	    int foodnum;
		  	    scanf("%d",&foodnum);
		  	    if(!map[foodnum][F+i])
				{
				     add(foodnum,F+i,1);
				     map[foodnum][F+i]=true;
				}
		  }
		  for(int j=1;j<=d;j++)
		  {
		  	    int drinknum;
		  	    scanf("%d",&drinknum);
		  	    if(!map[F+N+i][F+2*N+drinknum])
				{
				     add(F+N+i,F+2*N+drinknum,1);
				     map[F+N+i][F+2*N+drinknum]=true;
				}
		  }
	}
	MaxFlow();
	printf("%d\n",ans);
	return 0;
}

来源：csyzcyj