題干：

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

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.

題目大意：

有F種食物和D種飲料，每種食物或飲料只能供一頭牛享用，且每頭牛只享用一種食物和一種飲料。現(xiàn)在有N頭牛，每頭牛都有自己喜歡的食物種類(lèi)列表和飲料種類(lèi)列表，問(wèn)最多能使幾頭牛同時(shí)享用到自己喜歡的食物和飲料。（1 <= F <= 100, 1 <= D <= 100, 1 <= N <= 100）

解題報(bào)告：
以如下的順序依次建圖, 每條邊權(quán)值均為1
源點(diǎn) -> 食物 -> 牛 -> 牛 -> 飲料 -> 匯點(diǎn)

因?yàn)槊總€(gè)牛只能被選擇一次，所以需要將牛拆成兩個(gè)點(diǎn)并且連一條流量為1的邊。

AC代碼：

#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];//一共多少個(gè)點(diǎn)跑bfs，dis數(shù)組和q數(shù)組就開(kāi)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為源點(diǎn)到這個(gè)點(diǎn)的路徑上的最小邊權(quán) 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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