漂亮小姐姐点击就送:https://www.luogu.org/problemnew/show/P2891

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).

有F种食物和D种饮料,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。现在有n头牛,每头牛都有自己喜欢的食物种类列表和饮料种类列表,问最多能使几头牛同时享用到自己喜欢的食物和饮料。(1 <= f <= 100, 1 <= d <= 100, 1 <= n <= 100)

输入输出格式

输入格式:

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.

输出格式:

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

输入输出样例

输入样例#1: 复制

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
输出样例#1: 复制

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.

// luogu-judger-enable-o2
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std; const int N=1e4+;
const int M=3e4+;
const int INF=0x7fffffff; int n,F,D,S,T;
int head[N],num_edge;
struct Edge
{
int v,flow,nxt;
}edge[M<<]; inline int read()
{
char c=getchar();int num=,f=;
for(;!isdigit(c);c=getchar())
f=c=='-'?-:f;
for(;isdigit(c);c=getchar())
num=num*+c-'';
return num*f;
} inline void add_edge(int u,int v,int flow)
{
edge[++num_edge].v=v;
edge[num_edge].flow=flow;
edge[num_edge].nxt=head[u];
head[u]=num_edge;
} int dep[N];
inline bool bfs()
{
memset(dep,,sizeof(dep));
queue<int> que;
que.push(S),dep[S]=;
int now,v;
while(!que.empty())
{
now=que.front(),que.pop();
for(int i=head[now];i;i=edge[i].nxt)
{
if(edge[i].flow)
{
v=edge[i].v;
if(dep[v])
continue;
dep[v]=dep[now]+;
if(v==T)
return ;
que.push(v);
}
}
}
return ;
} int dfs(int now,int flow)
{
if(now==T)
return flow;
int outflow=,tmp,v;
for(int i=head[now];i;i=edge[i].nxt)
{
if(edge[i].flow)
{
v=edge[i].v;
if(dep[v]!=dep[now]+)
continue;
tmp=dfs(v,min(flow,edge[i].flow));
if(tmp)
{
outflow+=tmp;
flow-=tmp;
edge[i].flow-=tmp;
edge[i^].flow+=tmp;
if(!flow)
return outflow;
}
}
}
dep[now]=;
return outflow;
} int main()
{
num_edge=;
n=read(),F=read(),D=read();
T=F+D+n*+;
for(int i=;i<=n;++i)
{
add_edge(i+F,i+F+n,);
add_edge(i+F+n,i+F,);
}
for(int i=;i<=F;++i)
{
add_edge(S,i,);
add_edge(i,S,);
}
for(int i=;i<=D;++i)
{
add_edge(i+F+*n,T,);
add_edge(T,i+F+n*,);
}
for(int i=,a,b,c;i<=n;++i)
{
b=read(),c=read();
while(b--)
{
a=read();
add_edge(a,i+F,);
add_edge(i+F,a,);
}
while(c--)
{
a=read();
add_edge(i+F+n,a+F+n*,);
add_edge(a+F+n*,i+F+n,);
}
}
int flow=;
while(bfs())
flow+=dfs(S,INF);
printf("%d",flow);
return ;
}

P2891 [USACO07OPEN]吃饭Dining的更多相关文章

  1. P2891 [USACO07OPEN]吃饭Dining(最大流+拆点)

    题目描述 Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she w ...

  2. P2891 [USACO07OPEN]吃饭Dining 最大流

    \(\color{#0066ff}{ 题目描述 }\) 有F种食物和D种饮料,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.现在有n头牛,每头牛都有自己喜欢的食物种类列表和饮料种类 ...

  3. 洛谷P2891 [USACO07OPEN]吃饭Dining

    题目描述 Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she w ...

  4. 洛谷 P2891 [USACO07OPEN]吃饭Dining

    裸的最大流. #include <cstdio> #include <cstring> #include <queue> const int MAXN = 4e3 ...

  5. 「洛谷P2891」[USACO07OPEN]吃饭Dining 解题报告

    P2891 [USACO07OPEN]吃饭Dining 题目描述 Cows are such finicky eaters. Each cow has a preference for certain ...

  6. [Luogu P2891/POJ 3281/USACO07OPEN ]吃饭Dining

    传送门:https://www.luogu.org/problemnew/show/P2891 题面 \ Solution 网络流 先引用一句真理:网络流最重要的就是建模 今天这道题让我深有体会 首先 ...

  7. [USACO07OPEN]吃饭Dining

    嘟嘟嘟 这应该是网络流入门题之一了,跟教辅的组成这道题很像. 把每一只牛看成书,然后对牛拆点,因为每一只牛只要一份,食物和饮料分别看成练习册和答案. #include<cstdio> #i ...

  8. bzoj1711[USACO07OPEN]吃饭Dining

    题意 有F种食物和D种饮料,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.现在有n头牛,每头牛都有自己喜欢的食物种类列表和饮料种类列表,问最多能使几头牛同时享用到自己喜欢的食物和饮 ...

  9. BZOJ 1711 吃饭dining/Luogu P1402 酒店之王 拆点+最大流流匹配

    题意: (吃饭dining)有F种食物和D种饮料,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.现在有n头牛,每头牛都有自己喜欢的食物种类列表和饮料种类列表,问最多能使几头牛同时享 ...

随机推荐

  1. go语言的安装及环境配置

    Go语言开发环境搭建(ubuntu) 1.清理.卸载之前的 go 语言环境: 删除go目录:sudo rm -rf /usr/local/go 删除软链接(如果建立了软链接):sudo rm -rf ...

  2. 15. Scala并发编程模型Akka

    15.1 Akka介绍 1) Akka是Java虚拟机JVM平台上构建高并发.分布式和容错应用的工具包和运行时,可以理解成Akka是编写并发程序的框架 2) Akka用Scala语言写成,同时提供了S ...

  3. 简单添加自己的驱动程序到Linux内核树中

    背景: 移植4g模块的时候,看到文档中有添加驱动到内核的步骤,于是趁着这个机会,展开有关的学习. 了解更多可以访问:<Kconfig语法简介> Target :hi3531d Linux ...

  4. hdu 6625 three array (01-trie)

    大意: 给两个数组$a,b$, 要求重排使得$c_i=a_i\oplus b_i$字典序最小. 字典树上贪心取$n$次, 然后排序, 还不知道怎么证. #include <iostream> ...

  5. 用chattr命令防止系统中某个关键文件被修改

    用chattr命令防止系统中某个关键文件被修改:# chattr +i /etc/resolv.conf

  6. table固定宽度与自动宽度

    table-layout:auto(创建的table默认是此布局模式): 对table和td.th指定的宽度无效,浏览器会计算所有单元格的内容宽度才能得出一列宽度 如果想对单元格的内容自动折行需使用w ...

  7. ORA-01790 错误处理 SQL同一数据库中,两个查询结果数据类型不同时的union all 合

    转自: 出现这种错误,要先看一下是不是sql中有用到连接:union,unio all之类的,如果有,需要注意相同名称字段的数据类型一定要相同. 所以在union 或者union all 的时候造成了 ...

  8. ArcGIS Runtime SDK for Android 定位权限(GPS定位\网络定位)

    ACCESS_COARSE_LOCATION和ACCESS_FINE_LOCATION: android.permission.ACCESS_COARSE_LOCATION:是基站定位,即基于无线网络 ...

  9. javascript_05-操作符

    一元运算符 a++和++a //5 2 3 var a =1; var b = ++a + ++a; console.log(b) //4 1 3 var a =1; var b = a++ + ++ ...

  10. iview DatePicker 只能选本月

    html <FormItem label="活动时间" prop="activity_time"> <DatePicker v-model=& ...