【POJ - 3281】Dining（拆点建图，网络流最大流）

题干：

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 Fiintegers 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

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.

题目大意：

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料。（1 <= F <= 100, 1 <= D <= 100, 1 <= N <= 100）

解题报告：
以如下的顺序依次建图, 每条边权值均为1
源点 -> 食物 -> 牛 -> 牛 -> 饮料 -> 汇点

因为每个牛只能被选择一次，所以需要将牛拆成两个点并且连一条流量为1的边。

AC代码：

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
int n,F,D;
int tot;
struct Edge {int to,ne,w;
} e[100005 * 2];
int head[10005];
int st,ed;
int dis[10050],q[10005];//一共多少个点跑bfs，dis数组和q数组就开多大。
void add(int u,int v,int w) {e[++tot].to=v;e[tot].w=w;e[tot].ne=head[u];head[u]=tot;
}
bool bfs(int st,int ed) {memset(dis,-1,sizeof(dis));int front=0,tail=0;q[tail++]=st;dis[st]=0;while(front<tail) {int cur = q[front];if(cur == ed) return 1;front++;for(int i = head[cur]; i!=-1; i = e[i].ne) {if(e[i].w&&dis[e[i].to]<0) {q[tail++]=e[i].to;dis[e[i].to]=dis[cur]+1;}}}if(dis[ed]==-1) return 0;return 1;
}
int dfs(int cur,int limit) {//limit为源点到这个点的路径上的最小边权 if(limit==0||cur==ed) return limit;int w,flow=0;for(int i = head[cur]; i!=-1; i = e[i].ne) {        if(e[i].w&&dis[e[i].to]==dis[cur]+1) {w=dfs(e[i].to,min(limit,e[i].w));e[i].w-=w;e[i^1].w+=w;flow+=w;limit-=w;if(limit==0) break;}}if(!flow) dis[cur]=-1;return flow;
}
int dinic() {int ans = 0;while(bfs(st,ed)) ans+=dfs(st,0x7fffffff);return ans;
}
int main()
{cin>>n>>F>>D;int N = F + D + 2*n + 2;st = F + D + 2*n + 1;ed = F + D + 2*n + 2;tot=1;for(int i = 1; i<=ed; i++) head[i] = -1;for(int i = 1; i<=F; i++) add(st,i,1),add(i,st,0);for(int i = 1; i<=D; i++) add(F+2*n+i,ed,1),add(ed,F+2*n+i,0);for(int i = F+1; i<=F+n;i++) add(i,i+n,1),add(i+n,i,0);for(int numx,numy,i = 1; i<=n; i++) {scanf("%d%d",&numx,&numy);for(int x,j = 1; j<=numx; j++) scanf("%d",&x),add(x,F+i,1),add(F+i,x,0);for(int x,j = 1; j<=numy; j++) scanf("%d",&x),add(F+i+n,F+2*n+x,1),add(F+2*n+x,F+i+n,0);}printf("%d\n",dinic());return 0;
}

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