飞行员配对方案问题 网络流||二分图匹配

简单的二分图模型复习下最大流和二分图匹配

然而这题的数据没有SJ测不了,只能测前面的匹配数对不对

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<queue>
#define LL long long
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define efo(i,x) for(int i=last[x];i!=0;i=e[i].next)
using namespace std;
inline LL read()
{
LL d=0,f=1;char s=getchar();
while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
while(s>='0'&&s<='9'){d=d*10+s-'0';s=getchar();}
return d*f;
}
#define N 107
#define S 0
#define T 106
#define inf 100000000
struct edge
{
int f,y,next;
}e[N*N*2];
struct ansss
{
int x,y;
}ans
;
int anss=0;
int last
,ne=1;
int nm,n;

void add(int x,int y,int f)
{
e[++ne].f=f;e[ne].y=y;e[ne].next=last[x];last[x]=ne;
}
void add2(int x,int y)
{
add(x,y,1);add(y,x,0);
}

queue<int>q;
int h[N+5];
bool bfs(int s,int t)
{
memset(h,-1,sizeof(h));
q.push(s);h[s]=1;
while(!q.empty())
{
int now=q.front();
q.pop();
efo(i,now)
if(e[i].f&&h[e[i].y]==-1)
{
h[e[i].y]=h[now]+1;
q.push(e[i].y);
}
}
if(h[t]==-1)return 0;
else return 1;
}

int dfs(int k,int t,int maxf)
{
if(k==t)return maxf;
int f,ret=0;
efo(i,k)
if(e[i].f&&h[e[i].y]==h[k]+1)
{
f=dfs(e[i].y,t,min(maxf-ret,e[i].f));
e[i].f-=f;
e[i^1].f+=f;
ret+=f;
if(ret==maxf)return ret;
}
return ret;
}

int dinic(int s,int t)
{
int ret=0;
while(bfs(s,t))ret+=dfs(s,t,inf);
return ret;
}

bool getans(int k,int pre)
{
if(k==S)return 1;
efo(i,k)
if(e[i].f&&(e[i].y<=nm||k==T))
{
if(getans(e[i].y,k))
{
if(k!=S&&k!=T)
{
if(pre!=S&&pre!=T)
{
ans[++anss].x=k;
ans[anss].y=pre;
}
}
}
}
return 0;
}

bool anscom(ansss a,ansss b)
{
return a.x<b.x;
}

int main()
{
nm=read(),n=read();
int xx=read(),yy=read();
while(xx+yy!=-2)
{
add2(xx,yy);
xx=read(),yy=read();
}
fo(i,1,nm)add2(S,i);
fo(i,nm+1,n)add2(i,T);
cout<<endl<<endl<<dinic(S,T)<<endl;
getans(T,-1);
sort(ans+1,ans+anss+1,anscom);
fo(i,1,anss)
cout<<ans[i].x<<' '<<ans[i].y<<endl;
return 0;
}

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<bitset>
#define LL long long
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define down(i,a,b) for(int i=a;i>=b;i--)
#define efo(i,x) for(int i=last[x];i!=0;i=e[i].next)
using namespace std;
inline LL read()
{
LL d=0,f=1;char s=getchar();
while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
while(s>='0'&&s<='9'){d=d*10+s-'0';s=getchar();}
return d*f;
}
#define N 105
#define M 105
struct edge
{
int y,next;
}e[N*M];
int last
,ne=0;
int n,nm,m,ans=0;
int l
,r
;
bitset<N>y;

void add(int x,int y)
{
e[++ne].y=y;e[ne].next=last[x];last[x]=ne;
}

bool match(int x)
{
efo(i,x)
if(y[e[i].y]==0)
{
y[e[i].y]=1;
if(l[e[i].y]==0||match(l[e[i].y]))
{
l[e[i].y]=x;
r[x]=e[i].y;
return 1;
}
}
return 0;
}

int main()
{
nm=read(),n=read();
int x=read(),yy=read();
while(x+yy!=-2)
{
add(x,yy-nm);
x=read(),yy=read();
}
memset(l,0,sizeof(l));
memset(r,0,sizeof(r));
fo(i,1,nm)
{
y.reset();
if(match(i))ans++;
}
cout<<ans<<endl;
fo(i,1,nm)
if(r[i])
{
cout<<i<<' '<<r[i]+nm<<endl;
}
return 0;
}