寒假Day17-POJ3281-Dining（最大流+多对一的匹配）

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

3

Hint

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.

 

 

 

AC代码：（不好写）bfs+dfs+dinic

#include<string.h>
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<cmath>
using namespace std;
#define inf 0x3f3f3f3f
//const int N=20020;
const int N=1100;

typedef long long ll;

int n,m,s,t,tot;
int head[N],nextt[N],dep[N],cur[N];
int f[N][N],d[N][N],e[N][N];

int bfs()//重新建图、进行分层建图
{
    queue<int>Q;
    memset(dep,-1,sizeof(dep));//这里注意一下是mem还是for清空，小心超时
    dep[s]=1;//dep[s]=1; 相当于给源点分层为1
    Q.push(s);//Q.push(s);
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        for(int i=1; i<=t; i++)
        {
            //如果可以到达且还没有访问则入队
            //可以到达的条件是剩余容量大于0
            //没有访问的条件是当前层数==-1
            if(e[u][i]>0&&dep[i]==-1)
            {
                dep[i]=dep[u]+1;
                Q.push(i);
            }
        }
    }
    return dep[t]!=-1;
}

int dfs(int u,int minflow)//u起点(第一次传进来是1,之后会变)/查找路径上的最小流量
{
    if(u==t)
        return minflow;
    int x;
    for(int v=1; v<=t; v++)
    {
        if(e[u][v]>0&&dep[v]==dep[u]+1&&(x=dfs(v,min(minflow,e[u][v]))))
            // if(e[u][v]>0&&dep[v]==dep[u]+1)

        {
            // x=dfs(v,min(minflow,e[u][v]));
            e[u][v]-=x;
            e[v][u]+=x;
            return x;
        }
    }
    return 0;//一定要写！！！
}



int dinic()
{
    int sum=0;
    while(bfs())//先进行分层
    {
        while(1)
        {
            int w=dfs(s,inf);
            if(w==0)
                break;//找不到增广路了
            sum+=w;//找到则进行累加
        }
    }
    return sum;
}

int main()
{
    int F,D;
    while(~scanf("%d %d %d",&n,&F,&D))
    {
        memset(e,0,sizeof(nextt));
        for(int i=1; i<=n; i++)
        {
            int ff,dd;
            scanf("%d %d",&ff,&dd);
            for(int j=1; j<=ff; j++)
                scanf("%d",&f[i][j]);
            for(int j=1; j<=dd; j++)
                scanf("%d",&d[i][j]);
        }
        //s-f、f-牛、牛-牛、牛-d、d-t
        //0 4 3 2 -> 0 1-4 5-7 8-9
        //s:0           f:1-F           牛:F+1,F+n
        //牛:F+n+1,F+2n d:F+2n+1,F+2n+D  t:F+2n+D+1

        s=0,t=F+2*n+D+1;
        for(int i=1; i<=F; i++) //s-f
            //add(s,i,1);//inf??
            e[s][i]=1;
        for(int i=1; i<=n; i++) //枚举牛,可以少些一些for循环
        {
            for(int j=1; j<=F; j++)
            {
                if(f[i][j])
                    // add(j,i+F,1);
                    // add(f[i][j],i+F,1);
                    e[f[i][j]][i+F]=1;
            }
            //add(F+i,F+n+j,1);
            e[F+i][F+n+i]=1;
            for(int j=1; j<=D; j++)
            {
                if(d[i][j])
                    e[F+n+i][F+2*n+d[i][j]]=1;
                //add(F+n+i,F+2*n+i,1);
                //add(F+n+i,F+2*n+d[i][j],1);
            }
        }
        for(int i=1; i<=D; i++)
            e[F+2*n+i][t]=1;
        //add(F+2*n+i,t,1);
        int ans=dinic();
        printf("%d\n",ans);
    }
    return 0;
}