题目大意:有n个人,已知每人有ki个糖纸,并且知道每张糖纸的颜色。其中,Bob希望能和同伴交换使得手上的糖纸数尽量多。他的同伴只会用手上的重复的交换手上没有的,并且他的同伴们之间不会产生交换。求出Bob能拥有的最大糖纸种数。

题目分析:对于Bob拥有的糖纸,从源点s连一条弧,容量为Bob拥有的数量;对于Bob的小伙伴,从Bob连一条弧向他拥有的糖纸,容量为拥有数量减1,对于他不拥有的糖纸,连一条有向弧从糖纸到他,容量为1;对于每一种糖纸,连一条弧向汇点t。最大流便是答案。

代码如下:

# include<iostream>
# include<cstdio>
# include<cmath>
# include<string>
# include<vector>
# include<list>
# include<set>
# include<map>
# include<queue>
# include<cstring>
# include<algorithm>
using namespace std; # define LL long long
# define REP(i,s,n) for(int i=s;i<n;++i)
# define CL(a,b) memset(a,b,sizeof(a))
# define CLL(a,b,n) fill(a,a+n,b) const double inf=1e30;
const int INF=1<<30;
const int N=80; struct Edge
{
int fr,to,cap,fw;
Edge(int _fr,int _to,int _cap,int _fw):fr(_fr),to(_to),cap(_cap),fw(_fw){}
};
vector<Edge>edges;
vector<int>G[N];
int vis[N],d[N],cur[N],s,t,n,m,mark[30]; void init()
{
s=0,t=n+m+1;
edges.clear();
REP(i,0,t+1) G[i].clear();
} void addEdge(int u,int v,int cap)
{
edges.push_back(Edge(u,v,cap,0));
edges.push_back(Edge(v,u,0,0));
int len=edges.size();
G[u].push_back(len-2);
G[v].push_back(len-1);
} bool BFS()
{
queue<int>q;
CL(vis,0);
d[s]=0;
vis[s]=1;
q.push(s);
while(!q.empty())
{
int x=q.front();
q.pop();
REP(i,0,G[x].size()){
Edge &e=edges[G[x][i]];
if(!vis[e.to]&&e.cap>e.fw){
vis[e.to]=1;
d[e.to]=d[x]+1;
q.push(e.to);
}
}
}
return vis[t];
} int DFS(int x,int a)
{
if(x==t||a==0) return a;
int flow=0,f;
for(int &i=cur[x];i<G[x].size();++i){
Edge &e=edges[G[x][i]];
if(d[e.to]==d[x]+1&&(f=DFS(e.to,min(a,e.cap-e.fw)))>0){
e.fw+=f;
edges[G[x][i]^1].fw-=f;
flow+=f;
a-=f;
if(a==0) break;
}
}
return flow;
} int Dinic()
{
int flow=0;
while(BFS()){
CL(cur,0);
flow+=DFS(s,INF);
}
return flow;
} int main()
{
int T,k,r,cas=0;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
init();
CL(mark,0);
scanf("%d",&k);
while(k--)
{
scanf("%d",&r);
++mark[r];
}
REP(i,1,m+1) if(mark[i]) addEdge(s,i,mark[i]);
REP(i,1,n){
CL(mark,0);
scanf("%d",&k);
while(k--){
scanf("%d",&r);
++mark[r];
}
REP(j,1,m+1){
if(mark[j]>=2) addEdge(m+i+1,j,mark[j]-1);
else if(mark[j]==0) addEdge(j,m+i+1,1);
}
}
REP(i,1,m+1) addEdge(i,t,1);
printf("Case #%d: %d\n",++cas,Dinic());
}
  return 0;
}

  

05-27 01:07