标签：

Transfer water

2 10 20 30
1 3 2
2 4 1
1 2
2 1 2
0 0 0 0

30

Hint
In  3‐dimensional  space  Manhattan  distance  of  point  A  (x1,  y1,  z1)  and  B(x2,  y2,  z2)  is |x2‐x1|+|y2‐y1|+|z2‐z1|. 
 

/* ***********************************************
Author        :CKboss
Created Time  :2015年07月06日 星期一 09时23分30秒
File Name     :HDOJ4009.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int maxn=1200;
const int INF=0x3f3f3f3f;

int n,X,Y,Z;

struct POS
{
	int a,b,c;
}pos[maxn];

struct Edge
{
	int u,v,cost;
}edge[maxn*maxn];

int en;
int pre[maxn],id[maxn],vis[maxn],in[maxn];

void init() { en=0; }

int zhuliu(int root,int n,int m,Edge edge[])
{
	int res=0,v;
	while(true)
	{
		for(int i=0;i<n;i++) in[i]=INF;
		for(int i=0;i<m;i++)
		{
			if(edge[i].u!=edge[i].v&&edge[i].cost<in[edge[i].v])
			{
				pre[edge[i].v]=edge[i].u;
				in[edge[i].v]=edge[i].cost;
			}
		}
		for(int i=0;i<n;i++)
		{
			if(i!=root&&in[i]==INF) return -1;
		}
		int tn=0;
		memset(id,-1,sizeof(id));
		memset(vis,-1,sizeof(vis));
		in[root]=0;
		for(int i=0;i<n;i++)
		{
			res+=in[i];
			v=i;
			while(vis[v]!=i&&id[v]==-1&&v!=root)
			{
				vis[v]=i; v=pre[v];
			}
			if(v!=root&&id[v]==-1)
			{
				for(int u=pre[v];u!=v;u=pre[u])
					id[u]=tn;
				id[v]=tn++;
			}
		}
		if(tn==0) break;
		for(int i=0;i<n;i++)
			if(id[i]==-1) id[i]=tn++;
		for(int i=0;i<m;)
		{
			v=edge[i].v;
			edge[i].u=id[edge[i].u];
			edge[i].v=id[edge[i].v];
			if(edge[i].u!=edge[i].v)
				edge[i++].cost-=in[v];
			else
				swap(edge[i],edge[--m]);
		}
		n=tn;
		root=id[root];
	}
	return res;
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	while(scanf("%d%d%d%d",&n,&X,&Y,&Z)!=EOF)
	{
		if(n==0&&X==0&&Y==0&&Z==0) break;

		init();

		for(int i=1,x,y,z;i<=n;i++)
		{
			scanf("%d%d%d",&x,&y,&z);
			pos[i]=(POS){x,y,z};
		}

		/// root 0 is water
		for(int i=1;i<=n;i++)
		{
			int hight = pos[i].c;
			edge[en++]=(Edge){0,i,hight*X};
		}

		for(int i=1,m;i<=n;i++)
		{
			int to,from=i;
			scanf("%d",&m);
			for(int j=0;j<m;j++)
			{
				scanf("%d",&to);
				if(from==to) continue;

				int dist = abs(pos[to].a-pos[from].a)+abs(pos[to].b-pos[from].b)+abs(pos[to].c-pos[from].c);
				int h_to = pos[to].c;
				int h_from = pos[from].c;

				if(h_from>=h_to)
				{
					edge[en++]=(Edge){from,to,dist*Y};
				}
				else
				{
					edge[en++]=(Edge){from,to,dist*Y+Z};
				}
			}
		}

		/// zhuliu
		int lens = zhuliu(0,n+1,en,edge);
		if(lens==-1) puts("poor XiaoA");
		else printf("%d\n",lens);
	}
    
    return 0;
}

HDOJ 4009 Transfer water 最小树形图

标签：

原文地址：http://blog.csdn.net/ck_boss/article/details/46771535