网络流24题 飞行员配对方案问题

题目链接：https://www.oj.swust.edu.cn/problem/show/1736

就是求最大匹配，用匈牙利算法就可以了。

如果不会匈牙利算法  推荐博客：https://www.cnblogs.com/PencilWang/p/5769385.html

用匈牙利算法的代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<vector>
#include<set>
#include<cstdio>
#include<string>
#include<deque> 
using namespace std;
typedef long long LL;
#define eps 1e-8
#define INF 0x3f3f3f3f
#define maxn 105
/*struct point{
    int u,w;
};
bool operator <(const point &s1,const point &s2)
{
    if(s1.w!=s2.w)
    return s1.w>s2.w;
    else
    return s1.u>s2.u;
}*/
int pre[maxn],vis[maxn],head[maxn];
int n,m,k,t,cnt,ans;
struct node{
    int v,next;
}edge[maxn*maxn]; 
void init()
{
    fill(head,head+maxn-1,-1);
    fill(pre,pre+maxn-1,-1);
    cnt=ans=0;
}
void add(int u,int v)
{
    edge[++cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=cnt;
}
bool hungry(int u)
{
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(vis[v])
        continue;
        vis[v]=1;
        if(pre[v]==-1||hungry(pre[v]))
        {
            pre[v]=u;
            return true;
        }
    }
    return false;
}
int main()
{
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        init();
        int u,v;
        while(scanf("%d%d",&u,&v)&&(u!=-1&&v!=-1))
        {
            add(v,u);
        }
        for(int i=m+1;i<=n;i++)
        {
            fill(vis,vis+maxn-1,0);
            if(hungry(i))
            ans++;
        }
        printf("%d
",ans);
        for(int i=1;i<=m;i++)
        {
            if(pre[i]!=-1)
            printf("%d %d
",i,pre[i]);
        }
    }
    return 0;
}

也要用最大流写一下：

不会最大流的话：推荐博客：https://blog.csdn.net/x_y_q_/article/details/51999466

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<vector>
#include<set>
#include<cstdio>
#include<string>
#include<deque> 
using namespace std;
typedef long long LL;
#define eps 1e-8
#define INF 0x3f3f3f
#define maxn 105
/*struct point{
    int u,w;
};
bool operator <(const point &s1,const point &s2)
{
    if(s1.w!=s2.w)
    return s1.w>s2.w;
    else
    return s1.u>s2.u;
}*/
queue<int>q;
int capacity[maxn][maxn],depth[maxn];//数组capacity记录弧的容量 
int n,m,k,t;
int BFS(int src,int des)//BFS分层 
{
    while(!q.empty())
    q.pop();
    memset(depth,-1,sizeof(depth));
    depth[src]=0;
    q.push(src);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=1;i<=n+1;i++)
        {
            if(depth[i]==-1&&capacity[u][i]>0)
            {
                depth[i]=depth[u]+1;
                q.push(i);
            }
        }
    }
    return depth[des]!=-1;
}
int DFS(int u,int min1)//DFS不断找增广路 
{
    if(u==n+1)
    return min1;
    for(int i=1;i<=n+1;i++)
    {
        if(depth[i]==depth[u]+1&&capacity[u][i]>0)
        {
            int increase=DFS(i,min(min1,capacity[u][i]));
            if(increase>0)
            {
                capacity[u][i]-=increase;
                capacity[i][u]+=increase;
                return increase;
            }
        }
    }
    return -1;
}
int maxflow(int src,int des)
{
    int sumflow=0,increase=0;
    while(BFS(src,des))
    {
        while((increase=DFS(src,INF))>0)
        {
            sumflow+=increase;
        }
    }
    return sumflow;
}
int main()
{
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        memset(capacity,0,sizeof(capacity));
        int u,v;
        while(scanf("%d%d",&u,&v)&&u!=-1&&v!=-1)
        {
            capacity[u][v]+=1;
        }
        for(int i=1;i<=m;i++)//添加点0作为源点，并且点1到点m这些点与点0的容量为1 
        capacity[0][i]+=1;
        for(int i=m+1;i<=n;i++)//添加点n+1作为汇点，。。。 
        capacity[i][n+1]+=1;
        printf("%d
",maxflow(0,n+1));
        for(int i=1;i<=m;i++)//这里因为要输出两两组合的数字，所以我们可以看哪些弧增加了反向边 
        {
            for(int j=m+1;j<=n;j++)
            {
                if(capacity[j][i]>0)//i与j之间增加了一条反向边 
                printf("%d %d
",i,j);
            }
         } 
    }
    return 0;
}