Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering problems, all the stalls in the new barn are different. For the first week, Farmer John randomly assigned cows to stalls, but it quickly became clear that any given cow was only willing to produce milk in certain stalls. For the last week, Farmer John has been collecting data on which cows are willing to produce milk in which stalls. A stall may be only assigned to one cow, and, of course, a cow may be only assigned to one stall.
Given the preferences of the cows, compute the maximum number of milk-producing assignments of cows to stalls that is possible.
Line 1: | One line with two integers, N (0 <= N <= 200) and M (0 <= M <= 200). N is the number of cows that Farmer John has and M is the number of stalls in the new barn. |
Line 2..N+1: | N lines, each corresponding to a single cow. The first integer (Si) on the line is the number of stalls that the cow is willing to produce milk in (0 <= Si<= M). The subsequent Si integers on that line are the stalls in which that cow is willing to produce milk. The stall numbers will be integers in the range (1..M), and no stall will be listed twice for a given cow. |
5 5 2 2 5 3 2 3 4 2 1 5 3 1 2 5 1 2
A single line with a single integer, the maximum number of milk-producing stall assignments that can be made.
4
题目:http://ace.delos.com/usacoprob2?a=embNcvzXart&S=stall4
题意:给你n头牛,和m个墙,每头牛有自己喜欢的墙,要求每堵墙只能有一头牛,求最多的匹配数
分析:典型的二分图匹配,加上虚拟源和汇,就变成最大流问题
代码:
/* ID: 15114582 PROG: stall4 LANG: C++ */ #include<cstdio> #include<iostream> #include<cstring> using namespace std; const int oo=1e9; const int mm=1e5; const int mn=444; int node,src,dest,edge; int ver[mm],flow[mm],next[mm]; int head[mn],work[mn],dis[mn],q[mn]; void prepare(int _node,int _src,int _dest) { node=_node,src=_src,dest=_dest; for(int i=0;i<node;++i)head[i]=-1; edge=0; } void addedge(int u,int v,int c) { ver[edge]=v,flow[edge]=c,next[edge]=head[u],head[u]=edge++; ver[edge]=u,flow[edge]=0,next[edge]=head[v],head[v]=edge++; } bool Dinic_bfs() { int i,u,v,l,r=0; for(i=0;i<node;++i)dis[i]=-1; dis[q[r++]=src]=0; for(l=0;l<r;++l) for(i=head[u=q[l]];i>=0;i=next[i]) if(flow[i]&&dis[v=ver[i]]<0) { dis[q[r++]=v]=dis[u]+1; if(v==dest)return 1; } return 0; } int Dinic_dfs(int u,int exp) { if(u==dest)return exp; for(int &i=work[u],v,tmp;i>=0;i=next[i]) if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0) { flow[i]-=tmp; flow[i^1]+=tmp; return tmp; } return 0; } int Dinic_flow() { int i,ret=0,delta; while(Dinic_bfs()) { for(i=0;i<node;++i)work[i]=head[i]; while(delta=Dinic_dfs(src,oo))ret+=delta; } return ret; } int main() { freopen("stall4.in","r",stdin); freopen("stall4.out","w",stdout); int n,m,u,v,c; while(~scanf("%d%d",&n,&m)) { prepare(n+m+2,0,n+m+1); for(u=1;u<=n;++u) { addedge(src,u,1); scanf("%d",&c); while(c--) { scanf("%d",&v); addedge(u,n+v,1); } } while(m)addedge(n+m--,dest,1); printf("%d\n",Dinic_flow()); } return 0; }