Transfer water
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 482 Accepted Submission(s): 139
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
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
Recommend
lcy
分析：这题认真看就会发现是最小树形图问题，之前
poj 3164 这题的那个算法没错，hdu这题wa的原因是数组开小了，不过改一下居然TLE。。。于是各种Debug，本地生成大数据，发现光排序就超时了。。。囧，再接着发现是stl的sort不给力阿，自己写了个qsort，1046ms飘过。。。之间还加上了读入优化
求200++ms写法。。。
代码：
#include<cstdio>
using namespace std;
const int mm=2222222;
const int mn=2222;
struct edge
{
int s,t,w;
}g[mm];
int a[mn],b[mn],c[mn];
int in[mn],p[mn],q[mn],mark[mn],w[mn],fp[mn],from[mn],vis[mn];
int i,j,k,n,x,y,z,e,ans,sum,r;
bool huan;
inline void addedge(int u,int v,int c)
{
g[e].s=u,g[e].t=v,g[e++].w=c;
}
inline int f(int a)
{
return a<0?-a:a;
}
inline int len(int i,int j)
{
return f(a[i]-a[j])+f(b[i]-b[j])+f(c[i]-c[j]);
}
inline void init(int& a)
{
char ch=getchar();
while (ch<'0'||ch>'9') ch=getchar();
for (a=0; ch>='0'&&ch<='9'; ch=getchar()) a=a*10+ch-48;
}
void swap(int i,int j)
{
edge tmp;
tmp=g[i],g[i]=g[j],g[j]=tmp;
}
void mysort(int l,int r)
{
int i=l,j=r,s=g[(l+r)>>1].t,w=g[(l+r)>>1].w;
while(i<=j)
{
while(g[i].t<s||(g[i].t==s&&g[i].w<w))++i;
while(g[j].t>s||(g[j].t==s&&g[j].w>w))--j;
if(i<=j)swap(i++,j--);
}
if(i<r)mysort(i,r);
if(l<j)mysort(l,j);
}
int main()
{
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
while(init(n),init(x),init(y),init(z),n+x+y+z)
{
for(i=1;i<=n;++i)init(a[i]),init(b[i]),init(c[i]);
for(e=0,i=1;i<=n;++i)
{
addedge(0,i,c[i]*x);
init(k);
while(k--)
{
init(j);
if(i==j)continue;
addedge(i,j,len(i,j)*y+(c[i]>=c[j]?0:z));
}
}
mysort(0,e-1);
for(i=0;i<=n;++i)fp[i]=p[i]=-1,in[i]=0,mark[i]=i;
for(i=0;i<e;++i)
if(p[g[i].t]<0)p[g[i].t]=i;
huan=1,ans=sum=0;
while(huan)
{
sum=huan=0;
for(i=1;i<=n;++i)
if(fp[j=mark[i]]>=0)
{
if(fp[i]<0)in[i]+=w[j],mark[i]=mark[mark[i]];
else
{
in[i]+=w[i],ans+=w[i];
if(g[++p[fp[i]]].t!=fp[i])p[fp[i]]=-1;
}
}
for(i=0;i<=n;++i)fp[i]=-1,vis[i]=0;
for(i=1;i<=n;++i)
if(p[i]>=0)
{
if(fp[j=mark[i]]<0||(fp[j]>=0&&w[j]>g[p[i]].w-in[i]))
w[j]=g[p[i]].w-in[i],fp[j]=i,from[j]=mark[g[p[i]].s];
}
for(i=1;i<=n;++i)
if(fp[i]>=0)sum+=w[i];
for(i=1;i<=n;++i)
if(!vis[i])
{
r=0,j=i;
while(j>0&&vis[j]>=0)
{
if(vis[j]>0)
{
huan=1;
while(q[--r]!=j)mark[q[r]]=j,vis[q[r]]=-1;
vis[j]=-1;
}
else if(!vis[j])vis[q[r++]=j]=1;
if(fp[j]>=0)j=from[j];
else j=-1;
}
while(r--)vis[q[r]]=fp[q[r]]=-1;
}
}
printf("%d\n",ans+sum);
}
return 0;
}