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[usaco]4.2.2偶图匹配 The Perfect Stall
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
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典型的偶图匹配问题
初始化是，把无向图转换成有向图，每一条无向边转化成从stall到cow的边。
之后进行匹配
如果cow和一个stall进行了匹配，那么把cow到stall的边转化成从cow到stall的边。
进行匹配的过程如下。
如果一个还没有匹配的cow和一个stall，之间有一条交互路径，那么把这条交互路径上的所有的边进行转向。。
同时加入这个cow和stall。
当搜索不到这样的路径时，搜索结束。
/*
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;
}
最后更新：2017-04-02 06:51:56