hdu 4009 Transfer water

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|.

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define inf 0x3f3f3f3f
using namespace std;
const int maxm=1200000;
const int maxn=1200;
struct node
{
    int x,y,z;
};
struct NN
{
    int u,v,w;
};
NN res[maxm];
int m;
node dd[1010];
int n,X,Y,Z;
int resCost,cnt;
bool ff;
int pre[maxn],id[maxn],vis[maxn],in[maxn];

void addEdge(int x,int y,int z)
{
    res[m].u=x;
    res[m].v=y;
    res[m].w=z;
    m++;
}

int myabs(int a)
{
    if(a<0)
    return -a;
    return a;
}

int directedMST(int root)
{
    int ans=0,nv=n,i;
    while (1)
    {
        for (i=0;i<nv;i++)
        {
            in[i]=inf;
        }
        for (i=0;i<m;i++)
        {
            int u=res[i].u;
            int v=res[i].v;
            if (res[i].w<in[v]&&u!=v)
            {
                pre[v]=u;
                in[v]=res[i].w;
            }
        }
        for (i=0;i<nv;i++)
        {
            if (i==root)continue;
            if (in[i]==inf)return -1;
        }
        int cntnode=0;
        memset(id,-1,sizeof(id[0])*(nv+3));
        memset(vis,-1,sizeof(vis[0])*(nv+3));
        in[root]=0;
        for (i=0;i<nv;i++)
        {
            ans+=in[i];
            int 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]=cntnode;
                }
                id[v]=cntnode++;
            }
        }
        if (cntnode==0)break;
        for (i=0;i<nv;i++)
        {
            if (id[i]==-1)id[i]=cntnode++;
        }
        for (i=0;i<m;i++)
        {
            int v=res[i].v;
            res[i].u=id[res[i].u];
            res[i].v=id[res[i].v];
            if (res[i].u!=res[i].v)
            {
                res[i].w-=in[v];
            }
        }
        nv=cntnode;
        root=id[root];
    }
    return ans;
}

int main()
{
    while(~scanf("%d%d%d%d",&n,&X,&Y,&Z))
    {
        if(n==0 && X==0 && Y==0 && Z==0)
        break;
        m=0;
        n++;
        for(int i=1;i<n;i++)
        {
            scanf("%d%d%d",&dd[i].x,&dd[i].y,&dd[i].z);
            addEdge(0,i,dd[i].z*X);
        }
        for(int i=1;i<n;i++)
        {
            int k;
            scanf("%d",&k);
            for(int j=0;j<k;j++)
            {
                int x;
                scanf("%d",&x);
                if(x==i)
                continue;
                if(dd[i].z>=dd[x].z)
                {
                    int cost=myabs(dd[i].x-dd[x].x)+myabs(dd[i].y-dd[x].y)+myabs(dd[i].z-dd[x].z);
                    addEdge(i,x,cost*Y);
                }
                else
                {
                    int cost=myabs(dd[i].x-dd[x].x)+myabs(dd[i].y-dd[x].y)+myabs(dd[i].z-dd[x].z);
                    addEdge(i,x,cost*Y+Z);
                }
            }
        }
        printf("%d
",directedMST(0));
    }
    return 0;
}

 有向图的最小生成树（最小树形图），更新模板，复杂度（O（nm））