HDU5518 : John's Fences

求出平面图的对偶图，那么需要选择一些环，使得这些环可以异或出所有环。

对于两个不同的区域，需要用一个代价最小的环把它们区分开，这对应最小割。

那么求出对偶图的最小割树，所有树边之和就是把所有区域都区分开的最小代价。

#include<cstdio>
#include<cmath>
#include<set>
#include<algorithm>
#include<map>
using namespace std;
typedef pair<int,int>PI;
typedef long long ll;
const int N=1010,M=2010,inf=~0U>>2;
int Case,cas,n,m,cnt,i,x,y,z;map<PI,int>T;
struct P{
  int x,y;
  P(){}
  P(int _x,int _y){x=_x,y=_y;}
  ll operator*(const P&b){return 1LL*x*b.y-1LL*y*b.x;}
}a[N];
struct E{
  int x,y,z;double o;
  E(){}
  E(int _x,int _y,int _z){x=_x,y=_y,z=_z,o=atan2(a[y].x-a[x].x,a[y].y-a[x].y);}
}e[M];
bool del[M];int from[M];
namespace GetArea{
struct cmp{bool operator()(int a,int b){return e[a].o<e[b].o;}};
set<int,cmp>g[N];set<int,cmp>::iterator k;int i,j,q[M],t;
void work(){
  for(i=0;i<m+m;i++)if(!del[i]){
    for(q[t=1]=j=i;;q[++t]=j=*k){
      k=g[e[j].y].find(j^1);k++;
      if(k==g[e[j].y].end())k=g[e[j].y].begin();
      if(*k==i)break;
    }
    ll s=0;
    for(j=1;j<=t;j++)s+=a[e[q[j]].x]*a[e[q[j]].y],del[q[j]]=1;
    if(s<=0)continue;
    for(cnt++,j=1;j<=t;j++)from[q[j]]=cnt;
  }
}
}
namespace GH{
struct E{int t,f;E*nxt,*pair;}*g[N],*d[N],pool[10000],*cur;
int n,m,i,e[M][3],S,T,h[N],gap[N],maxflow,vis[N],a[N],b[N],ans;
void init(int _n){n=_n;m=ans=0;}
inline void newedge(int x,int y,int z){e[++m][0]=x;e[m][1]=y;e[m][2]=z;}
inline void add(int s,int t,int f){
  E*p=cur++;p->t=t;p->f=f;p->nxt=g[s];g[s]=p;
  p=cur++;p->t=s;p->f=0;p->nxt=g[t];g[t]=p;
  g[s]->pair=g[t];g[t]->pair=g[s];
}
inline int min(int a,int b){return a<b?a:b;}
int sap(int v,int flow){
  if(v==T)return flow;
  int rec=0;
  for(E*p=d[v];p;p=p->nxt)if(h[v]==h[p->t]+1&&p->f){
    int ret=sap(p->t,min(flow-rec,p->f));
    p->f-=ret;p->pair->f+=ret;d[v]=p;
    if((rec+=ret)==flow)return flow;
  }
  if(!(--gap[h[v]]))h[S]=T;
  gap[++h[v]]++;d[v]=g[v];
  return rec;
}
void dfs(int x){
  vis[x]=1;
  for(E*p=g[x];p;p=p->nxt)if(p->f&&!vis[p->t])dfs(p->t);
}
void solve(int l,int r){
  if(l>=r)return;
  int i;
  for(cur=pool,i=1;i<=T;i++)g[i]=d[i]=NULL,h[i]=gap[i]=0;
  for(i=1;i<=m;i++)add(e[i][0],e[i][1],e[i][2]),add(e[i][1],e[i][0],e[i][2]);
  add(S,a[l],inf),add(a[r],T,inf);
  for(gap[maxflow=0]=T,i=1;i<=T;i++)d[i]=g[i],vis[i]=0;
  while(h[S]<T)maxflow+=sap(S,inf);
  ans+=maxflow;
  dfs(S);
  int L=l,R=r;
  for(i=l;i<=r;i++)if(vis[a[i]])b[L++]=a[i];else b[R--]=a[i];
  for(i=l;i<=r;i++)a[i]=b[i];
  solve(l,R),solve(L,r);
}
int work(){
  S=n+1;T=S+1;
  for(i=1;i<=n;i++)a[i]=i;
  solve(1,n);
  return ans;
}
}
inline int getid(){
  int x,y;
  scanf("%d%d",&x,&y);
  if(T[PI(x,y)])return T[PI(x,y)];
  T[PI(x,y)]=++n;
  a[n]=P(x,y);
  return n;
}
int main(){
  scanf("%d",&Case);
  for(cas=1;cas<=Case;cas++){
    n=cnt=0;
    T.clear();
    scanf("%d",&m);
    for(i=0;i<m;i++){
      x=getid();
      y=getid();
      scanf("%d",&z);
      e[i<<1]=E(x,y,z);
      e[i<<1|1]=E(y,x,z);
    }
    for(i=0;i<m+m;i++)del[i]=from[i]=0;
    for(i=1;i<=n;i++)GetArea::g[i].clear();
    for(i=0;i<m+m;i++)GetArea::g[e[i].x].insert(i);
    GetArea::work();
    GH::init(cnt+1);
    for(i=0;i<m+m;i+=2)GH::newedge(from[i]+1,from[i^1]+1,e[i].z);
    printf("Case #%d: %d\n",cas,GH::work());
  }
  return 0;
}

HDU5518 : John's Fences