标签:operator find set for bool iter str 区分 inline
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; }
标签:operator find set for bool iter str 区分 inline
原文地址:http://www.cnblogs.com/clrs97/p/7476662.html