[网络流24题]飞行员配对方案问题

题目描述

裸的二分图最大匹配，也可以用匈牙利算法解决。

#include<complex>
#include<cstdio>
using namespace std;
const int INF=0x3f3f3f3f;
const int N=1e3+7;
struct node{
    int v,f,nxt;
}e[N<<1];
int n,m,Enum=1,ans,s,t;
int front[N],cur[N],deep[N];
int q[N];
int qread()
{
    int x=0;
    char ch=getchar();
    while(ch<'0' || ch>'9')ch=getchar();
    while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x;
}
void Insert(int u,int v)
{
    e[++Enum].v=v;e[Enum].f=1;e[Enum].nxt=front[u];front[u]=Enum;
    e[++Enum].v=u;e[Enum].nxt=front[v];front[v]=Enum;
}
bool bfs()
{
    for(int i=0;i<=t;i++)
    {
        cur[i]=front[i];
        deep[i]=0;
    }
    int head=1,tail=0,u;
    deep[s]=1;
    q[++tail]=s;
    while(head<=tail)
    {
        u=q[head++];
        for(int i=front[u];i;i=e[i].nxt)
            if(e[i].f && !deep[e[i].v])
            {
                deep[e[i].v]=deep[u]+1;
                if(e[i].v==t)return 1;
                q[++tail]=e[i].v;
            }
    }
    return 0;
}
int dfs(int x,int cur_flow)
{
    if(x==t)return cur_flow;
    int rest=cur_flow,v;
    for(int &i=cur[x];i;i=e[i].nxt)
    {
        v=e[i].v;
        if(e[i].f && deep[v]==deep[x]+1 && rest)
        {
            int new_flow=dfs(v,min(e[i].f,rest));
            e[i].f-=new_flow;
            e[i^1].f+=new_flow;
            rest-=new_flow;
            if(!rest)return cur_flow;
        }
    }
    deep[x]=0;
    return cur_flow-rest;
}
void Dinic()
{
    while(bfs())
        ans+=dfs(s,INF);
    if(!ans)printf("No Solution!\n");
    else printf("%d\n",ans);
}
int main()
{
    scanf("%d%d",&n,&m);
    s=0;t=n+1;
    for(int i=1;i<=m;i++)
        Insert(s,i);
    for(int i=m+1;i<=n;i++)
        Insert(i,t);
    int u,v;
    while(scanf("%d%d",&u,&v)!=EOF)
        Insert(u,v);
    Dinic();
    return 0;
}