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Transfer water
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 4730 Accepted Submission(s): 1677
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|.
题意：n,X,Y,Z分别代表，n个村子，每个村子可以选择自己凿井或者从有水的村子里引水过来。如果选择自己凿井，那么所需要的费用就是自家高度（z坐标）乘以X
如果选择从别人家引水，那么费用是两个村子的欧几里得距离诚意Y，如果终点比起点要高那么还需要加上Z。 最后问你让所有人家里都有水，最少需要多少费用。
思路：创造一个超级源点0号点代表自家凿井，这道题其实不存在找不到路径的情况，因为最起码可以每个村子都自己凿井，也就是从0号店到每个村子各一条边，这样的话就是一个最小树形图的问题了。
这里用了网上大牛的模板。 具体的实现就不多说了。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 1010
#define INF 100000000
struct Point
{
int x,y,z;
}p[N];
struct Edge
{
int u,v,w;
}e[N*N];
int m,in[N],vis[N],pre[N],id[N];
void addedge(int u,int v,int w)
{
e[m].u=u;
e[m].v=v;
e[m++].w=w;
}
int dist(Point a,Point b)
{
return abs(a.x-b.x)+abs(a.y-b.y)+abs(a.z-b.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 main()
{
int n,X,Y,Z;
int k,v;
while(scanf("%d %d %d %d",&n,&X,&Y,&Z)&&n)
{
m=0;
for(int i=1;i<=n;i++)
{
scanf("%d %d %d",&p[i].x,&p[i].y,&p[i].z);
addedge(0,i,p[i].z*X);
}
for(int i=1;i<=n;i++)
{
scanf("%d",&k);
for(int j=1;j<=k;j++)
{
scanf("%d",&v);
if(i==v) continue;
int w=dist(p[i],p[v])*Y;
if(p[v].z>p[i].z) w+=Z;
addedge(i,v,w);
}
}
int ans=Directed_MST(0,n+1,m);
if(ans==-1)
printf("poor XiaoA\n");
else
printf("%d\n",ans);
}
return 0;
}
模板：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 1010///根据实际情况改变
#define INF 100000000
struct Edge
{
int u,v,w;
}e[N*N];
int m,in[N],vis[N],pre[N],id[N];
void addedge(int u,int v,int w)
{
e[m].u=u;
e[m].v=v;
e[m++].w=w;
}
int Directed_MST(int root,int NV,int NE)///没给根的情况下要假设一个超级源点，到各边的距离sum是全部边的和+1.
///如果求出来的和等于-1或者大于等于2*sum就不满足，否则减去sum就是正解
{
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;
///if(u==root) cntroot=i;cntroot-题目给的边的条数就是根的编号
}
}
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;///最小树形图的长度
}