POJ 3281 Dining
来源:互联网 发布:数据分析师证书考试 编辑:程序博客网 时间:2024/06/05 15:11
Dining
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 17557 Accepted: 7822
Description
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 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.
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.
Source
USACO 2007 Open Gold
#include<iostream>#include<cstring>#include<cstdio>#include<queue>using namespace std;#define maxm 500000#define maxn 100000int S,T,F,D,N;int head[maxn],dep[maxn],tot=1;struct Edge{ int to,value,next;}e[maxm];void Add_Edge(int u,int v,int w){ e[++tot].to=v;e[tot].value=w; e[tot].next=head[u];head[u]=tot;}queue<int> q;bool BFS(){ memset(dep,-1,sizeof dep ); dep[S]=0;q.push(S); while(!q.empty()){ int u=q.front();q.pop(); for(int i=head[u];i;i=e[i].next){ int v=e[i].to; if(dep[v]==-1&&e[i].value){ dep[v]=dep[u]+1; q.push(v); } } } if(dep[T]!=-1) return true; else return false;}int DFS(int u,int flow){ int ret=0; if(u==T) return flow; for(int i=head[u];i;i=e[i].next){ int v=e[i].to; if(dep[v]==dep[u]+1&&e[i].value){ int x=DFS(v,min(e[i].value,flow-ret)); ret+=x;flow-=x; e[i].value-=x;e[i^1].value+=x; } } return ret;}void Dinic(){ int ans=0; while(BFS()) ans+=DFS(S,1e9); printf("%d\n",ans); return;}int main(){ scanf("%d%d%d",&N,&F,&D); T=N+N+F+D+20;S=0;int x; for(int i=1;i<=F;i++) Add_Edge(S,i,1),Add_Edge(i,S,0); for(int i=1,sumf,sumd;i<=N;i++){ scanf("%d%d",&sumf,&sumd); for(int j=1;j<=sumf;j++){ scanf("%d",&x);// Add_Edge(x,i+F,1); Add_Edge(i+F,x,0); } for(int x,j=1;j<=sumd;j++){ scanf("%d",&x); Add_Edge(i+F+N,N+N+F+x,1); Add_Edge(N+N+F+x,i+F+N,0); } } for(int i=1;i<=N;i++) Add_Edge(i+F,i+F+N,1),Add_Edge(i+F+N,i+F,0); for(int i=1;i<=D;i++) Add_Edge(i+F+N+N,T,1),Add_Edge(T,i+F+N+N,0); Dinic(); return 0;}
源点连边向食物,食物连边向牛,牛连边向饮料,饮料连边向汇点,跑Dinic,注意中间牛拆一下点不然可能会一头牛分配多个食物
- poj 3281 Dining Maxflow
- POJ 3281 Dining
- poj 3281 Dining //SAP
- POJ 3281 Dining
- poj 3281 Dining
- poj 3281 Dining
- POJ 3281 Dining
- POJ 3281 Dining
- poj 3281 Dining
- POJ-3281-Dining
- poj(3281)Dining
- POJ 3281 Dining
- poj 3281 Dining
- poj 3281 Dining
- POJ 3281 Dining
- POJ 3281 Dining
- POJ 3281 Dining dinic
- POJ 3281 — Dining
- 文件指针FILE*及文件描述符fd
- 标注 @JsonInclude(Include.NON_NULL)
- Mac OS tensorflow的环境搭建
- Spark源码学习笔记5-RpcEnv(Rpc抽象层)
- 模仿东京首页banner轮播,京东新闻上下滚动动画实现(动画实现)
- POJ 3281 Dining
- LeetCode[517] Super Washing Machines
- Extjs 组件查找
- 20170507@ 泛型进阶
- [FAQ05657][SAT]如何动态修改SIM卡应用名称
- C# 事件传值应用于窗口传值
- hadoop-ha之Hive
- android短信上送验证的实现及问题
- 【Java工具类】----正则表达式校验工具类