bzoj2960 跨平面 最小树形图 朱-刘算法

       首先用set找出所有的区域，然后对于一条边在相邻的区域之间连有向边，然后新建一个根连向所有点跑朱-刘算法即可。

AC代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<set>
#define ll long long
#define N 3005
#define M 30005
using namespace std;

int n,m,cnt,tot,fa[N],rch[N],id[N],mrk[N],q[M],blg[M]; bool bo[M];
struct point{ int x,y; }a[N];
struct edg{ int x,y,z; double k; }e[M];
struct cmp{ bool operator()(int x,int y){ return e[x].k<e[y].k; } };
ll crs(point u,point v){ return (ll)u.x*v.y-(ll)u.y*v.x; }
set<int,cmp>s[N];
int solve(int rt){
	int ans=0,i,j,k;
	while (1){
		for (i=0; i<=n; i++) rch[i]=1000000000;
		for (i=1; i<=tot; i++)
			if (e[i].z<rch[e[i].y] && e[i].x!=e[i].y){ rch[e[i].y]=e[i].z; fa[e[i].y]=e[i].x; }
		cnt=-1; rch[rt]=0;
		for (i=0; i<=n; i++) id[i]=mrk[i]=-1;
		for (i=0; i<=n; i++){
			ans+=rch[i];
			for (k=i; k!=rt && mrk[k]!=i && id[k]==-1; k=fa[k]) mrk[k]=i;
			if (k!=rt && id[k]==-1){
				id[k]=++cnt;
				for (j=fa[k]; j!=k; j=fa[j]) id[j]=cnt;
			}
		}
		if (cnt==-1) break;
		for (i=0; i<=n; i++) if (id[i]==-1) id[i]=++cnt;
		for (i=1; i<=tot; i++){
			k=rch[e[i].y];
			e[i].x=id[e[i].x]; e[i].y=id[e[i].y];
			if (e[i].x!=e[i].y) e[i].z-=k;
		}
		n=cnt; rt=id[rt];
	}
	return ans;
}
int main(){
	scanf("%d%d",&n,&m); int i,j,x;
	for (i=1; i<=n; i++) scanf("%d%d",&a[i].x,&a[i].y);
	for (i=1; i<=m; i++){
		scanf("%d%d%d",&e[i<<1].x,&e[i<<1].y,&e[i<<1].z);
		e[i<<1|1].x=e[i<<1].y; e[i<<1|1].y=e[i<<1].x;
		scanf("%d",&e[i<<1|1].z);
	}
	for (i=2; i<=(m<<1|1); i++){
		e[i].k=-atan2(a[e[i].y].y-a[e[i].x].y,a[e[i].y].x-a[e[i].x].x);
		s[e[i].x].insert(i);
	}
	memset(bo,1,sizeof(bo));
	set<int,cmp>::iterator it; ll tmp; int tp;
	for (i=2; i<=(m<<1|1); i++) if (bo[i]){
		tp=1; q[1]=i;
		while (1){
			bo[q[tp]]=0;
			it=s[e[q[tp]].y].find(q[tp]^1); it++;
			if (it==s[e[q[tp]].y].end()) it=s[e[q[tp]].y].begin();
			if (*it==i) break; else q[++tp]=*it;
		}
		tmp=0;
		for (j=1; j<=tp; j++) tmp+=crs(a[e[q[j]].x],a[e[q[j]].y]);
		if (tmp>=0){
			cnt++; for (j=1; j<=tp; j++) blg[q[j]]=cnt;
		}
	}
	x=1; n=cnt+1;
	for (i=2; i<=(m<<1|1); i++) if (e[i].z){
		tot++; x+=e[i].z; e[tot].z=e[i].z;
		e[tot].x=blg[i]; e[tot].y=blg[i^1];
	}
	for (i=0; i<n; i++){
		tot++; e[tot].x=n; e[tot].y=i; e[tot].z=x;
	}
	printf("%d\n",solve(n)-x);
	return 0;
}

by lych

2016.4.14