The Perfect Stall
Hal Burch
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.
PROGRAM NAME: stall4
INPUT FORMAT
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.
SAMPLE INPUT (file stall4.in)
5 5
2 2 5
3 2 3 4
2 1 5
3 1 2 5
1 2
OUTPUT FORMAT
A single line with a single integer, the maximum number of milk-producing stall assignments that can be made.
SAMPLE OUTPUT (file stall4.out)
4
-----------------------------------------------------
典型的偶圖匹配問題
初始化是，把無向圖轉換成有向圖，每一條無向邊轉化成從stall到cow的邊。
之後進行匹配
如果cow和一個stall進行了匹配，那麼把cow到stall的邊轉化成從cow到stall的邊。
進行匹配的過程如下。
如果一個還沒有匹配的cow和一個stall，之間有一條交互路徑，那麼把這條交互路徑上的所有的邊進行轉向。。
同時加入這個cow和stall。
當搜索不到這樣的路徑時，搜索結束。
/*
ID:yunleis2
PROG:stall4
LANG:C++
*/
#include<iostream>
#include<fstream>
using namespace std;
const int maxn=201;
const int maxm=201;
int n,m;
int metri[maxn][maxm];
bool cow[maxn];
int cownext[maxm];
int stallnext[maxn];
bool stall[maxm];
bool cowvisited[maxn];
bool search(int p);
int main()
{
fstream fin("stall4.in",ios::in );
fin>>n>>m;
for(int i=1;i<=n;i++){
int a,b;
fin>>a;
for(int j=0;j<a;j++){
fin>>b;
metri[i][b]=1;
}
}
while(true){
bool flag=false;
for(int s=1;s<=n;s++)
cowvisited[s]=false;
for(int i=1;i<=m;i++){
if(!stall[i]){
if(search(i))
{
flag=true;
stall[i]=true;
//--i;continue;
}
}
}
if(!flag)
break;
}
int result=0;
#if 0
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cout<<metri[i][j]<<" ";
}
cout<<endl;
}
#endif
for(int i=1;i<=n;i++){
if(cow[i])
result++;
}
fstream fout("stall4.out",ios::out);
fout<<result<<endl;
//system("pause");
}
bool search(int p){
//stall p;
for(int i=1;i<=n;i++){
if(cowvisited[i])
continue;
if(metri[i][p]==1){
if(!cow[i]){
cow[i]=true;
cownext[i]=p;
metri[i][p]=2;
cowvisited[i]=true;
return true;
}
else if(metri[i][cownext[i]]==2){
cowvisited[i]=true;
bool flag=search(cownext[i]);
if(flag){
metri[i][p]=2;
metri[i][cownext[i]]=1;
cownext[i]=p;
return flag;
}
}
}
}
return false;
}