2015
04-14

# Transfer water

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.

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.

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.

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

30

HintIn  3�\dimensional  space  Manhattan  distance  of  point  A  (x1,  y1,  z1)  and  B(x2,  y2,  z2)  is |x2�\x1|+|y2�\y1|+|z2�\z1|.


/*

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define M 1101
#define INF 1000000000
using namespace std;
struct edge
{
int u , v , w;
}e[M * M];
struct point
{
int x,y,z;
}home[M];
int pre[M],id[M],vis[M],in[M];
int n,m,X,Y,Z;
int Directed_MST(int root,int NV,int NE)
{
int ret = 0;
while(true){
//步骤1：找到最小边
for(int i = 0;i < NV;i ++)
in[i] = INF;
for(int i = 0;i < NE;i ++){
int u = e[i].u , v = e[i].v;
if(e[i].w < in[v] && u != v){
pre[v] = u;
in[v] = e[i].w;
}
}
for(int i = 0;i < NV;i ++){
if(i == root) continue;
if(in[i] == INF) return -1;//除了根节点以外有点没有入边，则根无法到达他
}
int cntnode = 0;
memset(id,-1,sizeof(id));
memset(vis,-1,sizeof(vis));
//找环
in[root] = 0;
for(int i = 0;i < NV;i ++){//标记每个环，编号
ret += 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;//无环
for(int i = 0;i < NV;i ++)
if(id[i] == -1){
id[i] = cntnode ++;
}
//步骤3：缩点，重新标记
for(int i = 0;i < NE;i ++){
int v = e[i].v;
e[i].u = id[e[i].u];
e[i].v = id[e[i].v];
if(e[i].u != e[i].v) e[i].w -= in[v];
}
NV = cntnode;
root = id[root];
}
return ret;
}
int dis(int a,int b)
{
return abs(home[a].x - home[b].x) + abs(home[a].y - home[b].y) + abs(home[a].z - home[b].z);
}
{
e[m].u = u , e[m].v = v,e[m].w = w;
m ++;
}
int main()
{
while(scanf("%d%d%d%d",&n,&X,&Y,&Z) != EOF && n){
m = 0;
for(int i = 1;i <= n;i ++){
scanf("%d%d%d",&home[i].x,&home[i].y,&home[i].z);
}
for(int i = 1;i <= n;i ++){
int num , u = i , v , w;
scanf("%d",&num);
for(int j = 0;j < num;j ++){
scanf("%d",&v);
if(v == i) continue;
w = dis(u,v) * Y;
if(home[v].z > home[u].z) w += Z;
}
}
int ans = Directed_MST(0,n + 1,m);
printf("%d\n",ans);
}
return 0;
}

,
1. 网站做得很好看，内容也多，全。前段时间在博客园里看到有人说：网页的好坏看字体。觉得微软雅黑的字体很好看，然后现在这个网站也用的这个字体！nice!