UVA-1345 - Jamie's Contact Groups(最大流+二分)

题意:

有n(n ≤ 1000)个人和m(m≤500)个组。一个人可能属于很多组。现在请你从某些组中去掉几个人，使得每个人只属于一个组，并使得人数最多的组中人员数目最小。

分析:看到最多的最小,就可以猜到是一个二分答案ans的题,最大流是没想到...囧.

这题用最大流做,若Ai属于Bi,则连一条Ai到Bi的边,边值为1,然后连超级源点S到Ai的边,边值为1,表示仅有一个Ai,然后对于超级源点T,连Bi到T的边,边值为ans,表示Bi组织最多有ans个人,求最大流,若最大流答案等于n,则表示在每组人限制ans个的情况下能够分配组员到各组,若不等于,则说明在每组人限制ans个的情况下,不能分配组员到各组;

// File Name: 1345.cpp
// Author: Zlbing
// Created Time: 2013/4/22 14:14:53

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 0x3f3f3f3f
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)

const int MAXN=2005;
int n,m;
struct Edge{
int from,to,cap,flow;
};
bool cmp(const Edge& a,const Edge& b){
return a.from < b.from || (a.from == b.from && a.to < b.to);
}
struct Dinic{
int n,m,s,t;
vector<Edge> edges;
vector<int> G[MAXN];
bool vis[MAXN];
int d[MAXN];
int cur[MAXN];
void init(int n){
this->n=n;
for(int i=0;i<=n;i++)G[i].clear();
edges.clear();
}
void AddEdge(int from,int to,int cap){
edges.push_back((Edge){from,to,cap,0});
edges.push_back((Edge){to,from,0,0});//当是无向图时，反向边容量也是cap,有向边时，反向边容量是0
m=edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool BFS(){
CL(vis,0);
queue<int> Q;
Q.push(s);
d[s]=0;
vis[s]=1;
while(!Q.empty()){
int x=Q.front();
Q.pop();
for(int i=0;i<G[x].size();i++){
Edge& e=edges[G[x][i]];
if(!vis[e.to]&&e.cap>e.flow){
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[x]+1==d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0){
e.flow+=f;
edges[G[x][i]^1].flow-=f;
flow+=f;
a-=f;
if(a==0)break;
}
}
return flow;
}
//当所求流量大于need时就退出，降低时间
int Maxflow(int s,int t,int need){
this->s=s;this->t=t;
int flow=0;
while(BFS()){
CL(cur,0);
flow+=DFS(s,INF);
if(flow>need)return flow;
}
return flow;
}
//最小割割边
vector<int> Mincut(){
BFS();
vector<int> ans;
for(int i=0;i<edges.size();i++){
Edge& e=edges[i];
if(vis[e.from]&&!vis[e.to]&&e.cap>0)ans.push_back(i);
}
return ans;
}
void Reduce(){
for(int i = 0; i < edges.size(); i++) edges[i].cap -= edges[i].flow;
}
void ClearFlow(){
for(int i = 0; i < edges.size(); i++) edges[i].flow = 0;
}
};
vector<int> T[MAXN];
Dinic solver;
bool solve(int mid)
{
solver.init(n+m);
int s=0,t=n+m+1;
REP(i,1,n)
{
solver.AddEdge(s,i,1);
REP(j,0,T[i].size())
{
solver.AddEdge(i,n+T[i][j]+1,1);
}
}
REP(i,1,m)
{
solver.AddEdge(n+i,t,mid);
}
int ans=solver.Maxflow(s,t,INF);
if(ans==n)return true;
else return false;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
if(n==0&&m==0)break;
REP(i,0,n)T[i].clear();
char ch[20];
REP(i,1,n)
{
scanf("%s",ch);
char c;
int a;
c=getchar();
//printf("aaacccc%caaa\n",c);
while(c!='\n')
{
scanf("%d",&a);
//printf("aa%d\n",a);
T[i].push_back(a);
c=getchar();
//printf("cccc%caa\n",c);
}
}
int L=0,R=n;
while(L<R)
{
int mid=(R+L)/2;
printf("mid--%d\n",mid);
if(solve(mid))
{
R=mid;
}
else{
L=mid+1;
}
}
printf("%d\n",L);
}
return 0;
}