HDU4009 Transfer water【最小树形图】【不定根】

Transfer water

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)

Total Submission(s): 3943    Accepted Submission(s): 1415

Problem Description

XiaoA lives in a village. Last year flood rained the village. So they decide to move the whole village to the mountain nearby this year. There is no spring in the mountain, so each household could only dig a well or build a water line from other household.
If the household decide to dig a well, the money for the well is the height of their house multiplies X dollar per meter. If the household decide to build a water line from other household, and if the height of which supply water is not lower than the one
which get water, the money of one water line is the Manhattan distance of the two households multiplies Y dollar per meter. Or if the height of which supply water is lower than the one which get water, a water pump is needed except the water line. Z dollar
should be paid for one water pump. In addition,therelation of the households must be considered. Some households may do not allow some other households build a water line from there house. Now given the 3‐dimensional position (a, b, c) of every household the
c of which means height, can you calculate the minimal money the whole village need so that every household has water, or tell the leader if it can’t be done.

Input

Multiple cases.

First line of each case contains 4 integers n (1<=n<=1000), the number of the households, X (1<=X<=1000), Y (1<=Y<=1000), Z (1<=Z<=1000).

Each of the next n lines contains 3 integers a, b, c means the position of the i‐th households, none of them will exceeded 1000.

Then next n lines describe the relation between the households. The n+i+1‐th line describes the relation of the i‐th household. The line will begin with an integer k, and the next k integers are the household numbers that can build a water line from the i‐th
household.

If n=X=Y=Z=0, the input ends, and no output for that.

Output

One integer in one line for each case, the minimal money the whole village need so that every household has water. If the plan does not exist, print “poor XiaoA” in one line.

Sample Input

2 10 20 30

1 3 2

2 4 1

1 2

2 1 2

0 0 0 0

Sample Output

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|.

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

题目大意：一个村庄被洪水摧毁了，整个村庄都要转移。但是山上没有泉水，每户家庭只能

在自家挖一个水井或是修一个水渠从别的家庭引水。如果要修井，则修井费用和房子所在海

拔高度有关，每米X元。如果从别人的家里引水，如果从高于自己家高度的人家里引水，费

用为每米Y元。如果从低于自己家高度的人家里引水，每条要多花费Z元。现在给你这个村庄

N个家庭房屋的坐标(a,b,c)和三种花费X,Y,Z。接着给你各家之间能单向修建引水沟渠的限制。

问：能使全村庄的人喝上水的总修建费用最低为多少。若不能，则输出"poor XiaoA"。

思路：其实就是给你一个有向图，求有向图的最小树形图是多少。但是根结点不确定。在这

里可以假设一个根结点，也就是虚根，让所有家庭都从虚根引水(其实就是每家都自己修井)，

所以问题肯定有解，最后就是朱刘算法(模板)直接求最小树形图。

朱刘算法参考我的另一篇博文：http://blog.csdn.net/lianai911/article/details/42242371

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int MAXN = 1010;
const int MAXM = 1000010;

struct Node
{
    int from;
    int to;
    int w;
};
Node Edges[MAXM];

struct Node1
{
    int x;
    int y;
    int z;
};
Node1 Point[MAXN];

int Dist(Node1 a,Node1 b)
{
    return abs(a.x-b.x) + abs(a.y-b.y) + abs(a.z-b.z);
}

int pre[MAXN],vis[MAXN],flag[MAXN],In[MAXN],sum;

int ZhuLiu(int root,int N,int M)
{
    sum = 0;
    while(true)
    {
        for(int i = 0; i < N; ++i)
            In[i] = INT_MAX;
        for(int i = 0; i < M; ++i)
        {
            int u = Edges[i].from;
            int v = Edges[i].to;
            if(Edges[i].w < In[v] && u != v)
            {
                pre[v] = u;
                In[v] = Edges[i].w;
            }
        }

        for(int i = 0; i < N; ++i)
        {
            if(i == root)
                continue;
            if(In[i] == INT_MAX)
                return -1;
        }
        int CntNode = 0;
        memset(flag,-1,sizeof(flag));
        memset(vis,-1,sizeof(vis));
        In[root] = 0;
        for(int i = 0; i < N; ++i)
        {
            sum += In[i];
            int v = i;
            while(vis[v] != i && flag[v] == -1 && v != root)
            {
                vis[v] = i;
                v = pre[v];
            }
            if(v != root && flag[v] == -1)
            {
                for(int u = pre[v]; u != v; u = pre[u])
                    flag[u] = CntNode;
                flag[v] = CntNode++;
            }
        }
        if(CntNode == 0)
            break;
        for(int i = 0; i < N; ++i)
        {
            if(flag[i] == -1)
                flag[i] = CntNode++;
        }
        for(int i = 0; i < M; ++i)
        {
            int v = Edges[i].to;
            Edges[i].from = flag[Edges[i].from];
            Edges[i].to = flag[Edges[i].to];
            if(Edges[i].from != Edges[i].to)
                Edges[i].w -= In[v];
        }
        N = CntNode;
        root = flag[root];
    }
    return sum;
}

int main()
{
    int N,M,X,Y,Z,k,d;
    while(~scanf("%d%d%d%d",&N,&X,&Y,&Z) && (N||X||Y||Z))
    {
        for(int i = 0; i < N; ++i)
            scanf("%d%d%d",&Point[i].x,&Point[i].y,&Point[i].z);
        M = 0;
        for(int i = 0; i < N; ++i)
        {
            scanf("%d",&k);
            while(k--)
            {
                scanf("%d",&d);
                d--;
                Edges[M].from = i;
                Edges[M].to = d;
                Edges[M].w = Dist(Point[i],Point[d])*Y;
                if(Point[d].z > Point[i].z)
                    Edges[M].w += Z;
                M++;
            }
        }
        for(int i = 0; i < N; ++i)
        {
            Edges[M].from = N;
            Edges[M].to = i;
            Edges[M++].w = Point[i].z*X;
        }
        printf("%d\n",ZhuLiu(N,N+1,M));
    }

    return 0;
}