Marriage Match II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2307 Accepted Submission(s): 792
Problem Description
Presumably, you all have known the question of stable marriage match. A girl will choose a boy; it is similar as the game of playing house we used to play when we are kids. What a happy time as so many friends playing together. And
it is normal that a fight or a quarrel breaks out, but we will still play together after that, because we are kids.
Now, there are 2n kids, n boys numbered from 1 to n, and n girls numbered from 1 to n. you know, ladies first. So, every girl can choose a boy first, with whom she has not quarreled, to make up a family. Besides, the girl X can also choose boy Z to be her boyfriend
when her friend, girl Y has not quarreled with him. Furthermore, the friendship is mutual, which means a and c are friends provided that a and b are friends and b and c are friend.
Once every girl finds their boyfriends they will start a new round of this game—marriage match. At the end of each round, every girl will start to find a new boyfriend, who she has not chosen before. So the game goes on and on.
Now, here is the question for you, how many rounds can these 2n kids totally play this game?
Input
There are several test cases. First is a integer T, means the number of test cases.
Each test case starts with three integer n, m and f in a line (3<=n<=100,0<m<n*n,0<=f<n). n means there are 2*n children, n girls(number from 1 to n) and n boys(number from 1 to n).
Then m lines follow. Each line contains two numbers a and b, means girl a and boy b had never quarreled with each other.
Then f lines follow. Each line contains two numbers c and d, means girl c and girl d are good friends.
Output
For each case, output a number in one line. The maximal number of Marriage Match the children can play.
Sample Input
1 4 5 2 1 1 2 3 3 2 4 2 4 4 1 4 2 3
Sample Output
2
这道题一开始我用匈牙利+删边做的，后来做网络流时看了一位大神的博客上的一段分析深有体会，又用最大流做了一遍。
链接：http://blog.csdn.net/qq564690377/article/details/7857983
这里我就只分析二分匹配了，其实很简单，就是根据并查集将该连的关系全都连起来，然后有匈牙利跑一遍，
每跑一遍就将当前节点（女孩）与其当前匹配的节点（男孩）关系删掉，再继续跑匈牙利，直到跑出的最大
匹配数小与n为止，此时跑的遍数就是最大轮数。
现上代码：
网络流：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
#define Min(a,b) a<b?a:b
#define inf 1000000000
#define maxn 20000
#define maxh 400
using namespace std;
typedef struct //前向星存图
{
int to,next,w;
}node;
typedef struct
{
int x,t;
}dep;
node E[maxn];
int head[maxh],headx[maxh],deep[maxh],cnt;
int set[maxh],map[maxh][maxh];
int n,m,f;
////////////////////////////////wangluoliu//////////////////////////////
void init()
{
memset(head,-1,sizeof(head));
cnt=0;
}
void add(int a,int b,int c)
{
E[cnt].to=b;
E[cnt].w=c;
E[cnt].next=head[a];
head[a]=cnt++;
E[cnt].to=a;
E[cnt].w=0;
E[cnt].next=head[b];
head[b]=cnt++;
}
int min(int x,int y)
{
return x<y?x:y;
}
int bfs(int s,int t,int n)
{
memset(deep,255,sizeof(deep));
queue<dep>Q;
dep fir,nex;
fir.x=s;
fir.t=0;
deep[s]=0;
Q.push(fir);
while(!Q.empty())
{
fir=Q.front();
Q.pop();
for(int i=head[fir.x];i+1;i=E[i].next)
{
nex.x=E[i].to;
nex.t=fir.t+1;
if(deep[nex.x]!=-1||!E[i].w)
continue;
deep[nex.x]=nex.t;
Q.push(nex);
}
}
for(int i=0;i<=n;i++)
headx[i]=head[i];
return deep[t]!=-1;
}
int dfs(int s,int t,int flow)
{
if(s==t)
return flow;
int newflow=0;
for(int i=headx[s];i+1;i=E[i].next)
{
headx[s]=i;
int to=E[i].to;
int w=E[i].w;
if(!w||deep[to]!=deep[s]+1)
continue;
int temp=dfs(to,t,min(w,flow-newflow));
newflow+=temp;
E[i].w-=temp;
E[i^1].w+=temp;
if(newflow==flow)
break;
}
if(!newflow)deep[s]=0;
return newflow;
}
int Dinic(int s,int t,int m)
{
int sum=0;
while(bfs(s,t,m))
{
sum+=dfs(s,t,inf);
}
return sum;
}
////////////////////////////////bingchaji///////////////////////////////////
int findx(int x)
{
if(set[x]!=x)
set[x]=findx(set[x]);
return set[x];
}
void fun(int x,int y)
{
x=findx(x);
y=findx(y);
if(x!=y)
set[x]=y;
}
//////////////////////////////////////jiantu////////////////////////////
void built(int mid)
{
init();//前向星初始化
int s=2*n+1,t=2*n+2;
for(int i=1;i<=n;i++)
{
add(s,i,mid);
add(i+n,t,mid);
for(int j=1;j<=n;j++)
{
if(map[i][j])
{
add(i,j+n,1);
}
}
}
}
////////////////////////////////////////////////////////////////////////
int main()
{
int T;
int a,b;
int l,r,mid;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&f);
for(int i=0;i<=n;i++)//初始化
{
set[i]=i;
for(int j=0;j<=n;j++)
map[i][j]=0;
}
for(int i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
map[a][b]=1;
}
for(int i=0;i<f;i++)
{
scanf("%d%d",&a,&b);
fun(a,b);
}
for(int i=1;i<=n;i++)
{
int xi=findx(i);
for(int j=1;j<=n;j++)
{
int yi=findx(j);
if(xi==yi&&i!=j)
{
for(int k=1;k<=n;k++)
{
if(map[j][k])
{
map[i][k]=1;
}
}
}
}
}
l=0,r=n+1;
while(l!=r-1)
{
// printf("xcwdf\n");
mid=(l+r)/2;
built(mid);
if(Dinic(2*n+1,2*n+2,2*n+2)==n*mid)
{
l=mid;
}
else
{
r=mid;
}
}
printf("%d\n",l);
}
return 0;
}
二分匹配：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define maxn 200
using namespace std;
int markd[maxn],markx[maxn],xm[maxn][maxn];
int set[maxn];
int n,m,f;
int findx(int x)
{
if(x!=set[x])
set[x]=findx(set[x]);
return set[x];
}
void fun(int x,int y)
{
x=findx(x);
y=findx(y);
if(x!=y)
{
set[y]=x;
}
}
int xyl(int x)
{
for(int i=1;i<=n;i++)
{
if(xm[x][i]&&!markd[i])
{
markd[i]=1;
if(markx[i]==-1||xyl(markx[i]))
{
markx[i]=x;
return 1;
}
}
}
return 0;
}
int main()
{
int T;
int a,b;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&f);
memset(xm,0,sizeof(xm));
for(int i=1;i<=n;i++)
set[i]=i;
for(int i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
xm[a][b]=1;
}
for(int i=1;i<=f;i++)
{
scanf("%d%d",&a,&b);
fun(a,b);
}
for(int i=1;i<=n;i++)
{
int xi=findx(i);
for(int j=1;j<=n;j++)
{
int yi=findx(j);
if(xi==yi&&i!=j)
{
for(int k=1;k<=n;k++)
{
if(xm[j][k])
xm[i][k]=1;
}
}
}
}
int cnt=0;
while(1)
{
memset(markx,255,sizeof(markx));
int sum=0;
for(int i=1;i<=n;i++)
{
memset(markd,0,sizeof(markd));
sum+=xyl(i);
}
if(sum<n)
break;
cnt++;
for(int i=1;i<=n;i++)
{
xm[markx[i]][i]=0; //删掉当前匹配边
}
}
printf("%d\n",cnt);
}
return 0;
}