Codefoces 277 E. Binary Tree on Plane

题目链接：http://codeforces.com/problemset/problem/277/E

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参考了这篇题解：http://blog.csdn.net/Sakai_Masato/article/details/50775315

没看出来是费用流啊...我好菜啊。

我写得有一点和上面这篇题解不同：在判断是否无解时我直接记录最大流，判断最大流是否等于$n-1$(似乎是等价的...)

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<vector>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 using namespace std;
 #define maxn 1010
 #define inf 0x7fffffff
 #define llg int
 #define sqr(_) ((_)*(_))
 #define yyj(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
 llg n,m;
 
 struct node
 {
     llg u,v,c,next;
     double w;
 };
 
 struct FLOW
 {
     llg cnt,maxflow,S,T,pre[maxn],N;
     llg head[maxn],dl[maxn*maxn];
     bool bj[maxn];
     double mincost,dis[maxn];
     node e[maxn*maxn];
 
     void init()
     {
         memset(head,-,sizeof(head));
         mincost=cnt=maxflow=;
         maxflow=; mincost=;
     }
 
     void link(llg u,llg v,double w,llg c)
     {
         e[cnt].u=u,e[cnt].v=v,e[cnt].w=w,e[cnt].c=c;
         e[cnt].next=head[u],head[u]=cnt++;
 
         e[cnt].u=v,e[cnt].v=u,e[cnt].w=-w,e[cnt].c=;
         e[cnt].next=head[v],head[v]=cnt++;
     }
 
     void updata()
     {
         llg f=inf;
         for (llg i=T;i!=S;i=e[pre[i]].u) f=min(f,e[pre[i]].c);
         maxflow+=f;
         for (llg i=T;i!=S;i=e[pre[i]].u)
         {
             e[pre[i]].c-=f;
             e[pre[i]^].c+=f;
             mincost+=(double)f*e[pre[i]].w;
         }
     }
 
     bool spfa()
     {
         llg u,v;
         double w;
         memset(pre,-,sizeof(pre));
         memset(bj,,sizeof(bj));
         for (llg i=;i<=N;i++) dis[i]=inf;
         llg l=,r=;
         dl[r]=S; bj[S]=; dis[S]=;
         do
         {
             bj[u=dl[++l]]=;
             for (llg i=head[u];i!=-;i=e[i].next)
             {
                 v=e[i].v,w=e[i].w;
                 if (e[i].c && dis[v]>dis[u]+w)
                 {
                     dis[v]=dis[u]+w;
                     pre[v]=i;
                     if (!bj[v]) bj[v]=,dl[++r]=v;
                 }
             }
         }while (l<r);
 
         if (pre[T]==-) return ;
         return ;
     }
 
     void work()
     {
         while (spfa())
             updata();
     }
 
 }G;
 
 inline int getint()
 {
     int w=,q=; char c=getchar();
     while((c<'' || c>'') && c!='-') c=getchar(); if(c=='-') q=,c=getchar();
     while (c>='' && c<='') w=w*+c-'', c=getchar(); return q ? -w : w;
 }
 
 struct POINT{double x,y;}po[maxn];
 
 bool cmp(const POINT&x,const POINT&y) {return x.y==y.y?x.x<y.x:x.y<y.y;}
 
 double dis(const POINT&x,const POINT&y) {return sqrt(sqr(x.x-y.x)+sqr(x.y-y.y));}
 
 int main()
 {
     yyj("flow");
     cin>>n;
     G.init();
     G.N=n*+;
     for (llg i=;i<=n;i++) scanf("%lf%lf",&po[i].x,&po[i].y);
     sort(po+,po+n+,cmp);
     reverse(po+,po++n);
     for (llg i=;i<=n;i++)
         for (llg j=i+;j<=n;j++)
             if (po[i].y>po[j].y)
                 G.link(i,j+n,dis(po[i],po[j]),(llg));
     G.S=*n+,G.T=*n+;
     for (llg i=;i<=n;i++)
     {
         G.link(G.S,i,,(llg));
         G.link(i+n,G.T,,(llg));
     }
     G.work();
     if (G.maxflow!=n-) {cout<<-;} else printf("%.9lf",G.mincost);
     return ;
 }