HDU 5639 Deletion 二分+网络流

题意：bc round 74 div1

分析：

考虑删掉的边的形态, 就是我们经常见到的环套树这种结构, 参考平时这种图给出的方法, 如果一个图的每个点的出边只有一条,

那么一定会构成环套树这种结构. 于是问题可以转化成, 给无向图的每条边定向, 使得出度最大点的出度最小 (每个点的出度大小对应了删的次数).

题解给出三种做法，对每条边定向，我采取第二种

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <queue>
#include <cmath>
using namespace std;
typedef long long LL;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
const int maxn=4e3+5;
struct Edge
{
    int from,to,cap,flow;
    Edge(int u,int v,int c,int d):from(u),to(v),cap(c),flow(d) {}
};
int x[maxn],y[maxn],n,m;
struct dinic
{
    int s,t;
    vector<Edge>edges;
    vector<int>G[maxn];
    int d[maxn];
    int cur[maxn];
    bool vis[maxn];
    void init()
    {
       edges.clear();
       for(int i=0;i<maxn;++i)
         G[i].clear();
    }
    bool bfs()
    {
        memset(vis,0,sizeof(vis));
        queue<int>q;
        q.push(s);
        d[s]=0;
        vis[s]=1;
        while(!q.empty())
        {
            int x=q.front();
            q.pop();
            for(int i=0; i<G[x].size(); i++)
            {
                Edge &e= edges[G[x][i]];
                if(!vis[e.to]&&e.cap>e.flow)
                {
                    vis[e.to]=1;
                    d[e.to]=d[x]+1;
                    q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int dfs(int x,int a)
    {
        if(x==t||a==0)return a;
        int flow=0,f;
        for(int &i=cur[x]; i<G[x].size(); i++)
        {
            Edge &e=edges[G[x][i]];
            if(d[x]+1==d[e.to]&&(f=dfs(e.to,min(a,e.cap-e.flow))))
            {
                e.flow+=f;
                edges[G[x][i]^1].flow-=f;
                flow+=f;
                a-=f;
                if(a==0)break;
            }
        }
        return flow;
    }
    int maxflow(int s,int t)
    {
        this->s=s;
        this->t=t;
        int flow=0;
        while(bfs())
        {
            memset(cur,0,sizeof(cur));
            flow+=dfs(s,INF);
        }
        return flow;
    }
    void addedge(int u,int v,int c)
    {
        Edge x(u,v,c,0),y(v,u,0,0);
        edges.push_back(x);
        edges.push_back(y);
        int l=edges.size();
        G[u].push_back(l-2);
        G[v].push_back(l-1);
    }
}solve;
bool fun(int s)
{
   solve.init();
   for(int i=1;i<=n;++i)
    solve.addedge(0,i,s);
   for(int i=1;i<=m;++i)
   {
    solve.addedge(x[i],i+n,1);
    solve.addedge(y[i],i+n,1);
   }
   for(int i=n+1;i<=n+m;++i)
   solve.addedge(i,n+m+1,1);
   int ans=solve.maxflow(0,n+m+1);
   if(ans==m)return 1;
   return 0;
}
int getans()
{
    if(m==0)return 0;
    int l=0,r=m;
    while(l<r)
    {
        int mid=(l+r)>>1;
        if(fun(mid))r=mid;
        else l=mid+1;
    }
    return (l+r)>>1;
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;++i)
         scanf("%d%d",&x[i],&y[i]);
        printf("%d
",getans());
    }
    return 0;
}

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