洛谷2756飞行员配对方案问题

题目链接：飞行员配对方案问题

很多人把这题当做二分图匹配的模板做，匈牙利的时间复杂度为\(O(nm)\)

但是如果用dinic去做时间就会是\(O(\sqrt nm)\)

建立超级源点\(s\)和\(t\)，直接按照输入连边，源点和汇点分别连向一个点集即可

#include<iostream>
#include<string>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<vector>
#include<queue>
#include<map>
using namespace std;
#define maxd 1e9+7
struct network_flows{
    struct node{
        int from,to,nxt,flow;
    }sq[100100];
    int all,dep[100100],head[100100],cur[100100],n,m,s,t;
    bool vis[100100];

   void init(int n,int m)
    {
        this->s=n+m+1;this->t=n+m+2;
        this->n=n+m+2;this->all=1;
        memset(head,0,sizeof(head));
    }

    void add(int u,int v,int w)
    {
        all++;sq[all].from=u;sq[all].to=v;sq[all].nxt=head[u];sq[all].flow=w;head[u]=all;
        all++;sq[all].from=v;sq[all].to=u;sq[all].nxt=head[v];sq[all].flow=0;head[v]=all;
    }

    bool bfs()
    {
        queue<int> q;int i;
        memset(vis,0,sizeof(vis));
        vis[s]=1;q.push(s);dep[s]=0;
        while (!q.empty())
        {
            int u=q.front();q.pop();
            for (i=head[u];i;i=sq[i].nxt)
            {
                int v=sq[i].to;
                if ((!vis[v]) && (sq[i].flow))
                {
                    vis[v]=1;dep[v]=dep[u]+1;q.push(v);
                }
            }
        }
        if (!vis[t]) return 0;
        for (i=1;i<=n;i++) cur[i]=head[i];
        return 1;
    }

    int dfs(int now,int to,int lim)
    {
        if ((!lim) || (now==to)) return lim;
        int i,sum=0;
        for (i=cur[now];i;i=sq[i].nxt)
        {
            cur[now]=i;int v=sq[i].to;
            if (dep[now]+1==dep[v])
            {
                int f=dfs(v,to,min(lim,sq[i].flow));
                if (f)
                {
                    lim-=f;sum+=f;
                    sq[i].flow-=f;
                    sq[i^1].flow+=f;
                    if (!lim) break;
                }
            }
        }
        return sum;
    }

    int work()
    {
        int ans=0;
        while (bfs()) ans+=dfs(s,t,maxd);
        return ans;
    }

    void out(int n,int m)
    {
        int i,j;
        for (i=1;i<=n;i++)
        {
            for (j=head[i];j;j=sq[j].nxt)
            {
                if ((sq[j].to>n) && (sq[j].to<=n+m) && (!sq[j].flow))
                    printf("%d %d\n",i,sq[j].to);
            }
        }
    }
}dinic;
int n,m,s,t;

int read()
{
    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

void init()
{
    n=read();m=read();int i,j,u,v;
    dinic.init(n,m);
    s=n+m+1;t=n+m+2;
    for (i=1;i<=n;i++) dinic.add(s,i,1);
    for (i=1;i<=m;i++) dinic.add(n+i,t,1);
    while ((scanf("%d%d",&u,&v)) && (u!=-1) && (t!=-1))
        dinic.add(u,v,1);
}

void work()
{
    int ans=dinic.work();
    if (!ans) {printf("No Solution!");return;}
    printf("%d\n",ans);
    dinic.out(n,m);
}

int main()
{
    init();
    work();
    return 0;
}