Hdu 4009 Transfer water【最小树形图】
Transfer water
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 4702 Accepted Submission(s): 1665
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.
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
30HintIn 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
最小树形图
题目描述的比较复杂，加重了构图的困难....因为确实不会做....
参照大神的构图方法，然后跑跑了一发模板（模板点这里），然后过了.....
/*
http://blog.csdn.net/liuke19950717
*/
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=1005;
struct node
{
int u,v;
int len;
}edge[maxn*maxn];
int pre[maxn],id[maxn],vis[maxn];
int in[maxn],tot;
int dir_mst(int root,int n,int m)
{
int ans=0;
while(1)
{
//先找出所有点的最小入边
/*memset(in,inf,sizeof(in)); //这样写会报错！*/
for(int i=0;i<n;++i)
{
in[i]=inf;
}
for(int i=0;i<m;++i)
{
int u=edge[i].u,v=edge[i].v;
if(edge[i].len<in[v]&&u!=v)
{
pre[v]=u;
in[v]=edge[i].len;
}
}
for(int i=0;i<n;++i)
{
if(i==root)
{
continue;
}
if(in[i]==inf)
{
return -1; //如果某点入度为零，必定找不到
}
}
//检查这些边是否构成了环
memset(id,-1,sizeof(id));
memset(vis,-1,sizeof(vis));
in[root]=0;
int cnt=0;
for(int i=0;i<n;++i)//标记环
{
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)//缩点
{
for(int u=pre[v];u!=v;u=pre[u])
{
id[u]=cnt;
}
id[v]=cnt++;
}
}
if(cnt==0)
{
break;//无环
}
for(int i=0;i<n;++i)
{
if(id[i]==-1)
{
id[i]=cnt++;
}
}
//建立新图
for(int i=0;i<m;++i)
{
int u=edge[i].u,v=edge[i].v;
edge[i].u=id[u];
edge[i].v=id[v];
if(id[u]!=id[v])
{
edge[i].len-=in[v];
}
/*
edge[i].u=id[e[i].u];
edge[i].v=id[e[i].v];
if(edge[i].u!=edge[i].v)
{
edge[i].len-=in[v];
}*/
}
n=cnt;
root=id[root];
}
return ans;
}
int x[maxn],y[maxn],z[maxn];
int dis(int i,int j)//！！！！！！！！！啊啊啊啊啊
{
return abs(x[i]-x[j])+abs(y[i]-y[j])+abs(z[i]-z[j]);
}
int main()
{
int X,Y,Z,n;
while(scanf("%d%d%d%d",&n,&X,&Y,&Z),n|X|Y|Z)
{
tot=0;
for(int i=1;i<=n;++i)
{
scanf("%d%d%d",&x[i],&y[i],&z[i]);
node tp={0,i,z[i]*X};
edge[tot++]=tp;
}
for(int i=1;i<=n;++i)
{
int k;
scanf("%d",&k);
while(k--)
{
int t;
scanf("%d",&t);
if(i==t)
{
continue;
}
if(z[i]>=z[t])
{
int d=dis(i,t);
node tp={i,t,d*Y};
edge[tot++]=tp;
}
else
{
int d=dis(i,t);
node tp={i,t,d*Y+Z};
edge[tot++]=tp;
}
}
}
int root=0;
int ans=dir_mst(0,n+1,tot);
if(ans==-1)
{
printf("poor XiaoA\n");
}
else
{
printf("%d\n",ans);
}
}
return 0;
}