HDU4009 Transfer water —— 最小树形图 + 不定根 + 超级点

题目链接：https://vjudge.net/problem/HDU-4009

Transfer water

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 5612    Accepted Submission(s): 1997

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|.

 

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

 

 

 

 

题解：

1.可知这题肯定有解，因为大不了每一户都自己挖口井。

2.题目其实就是要求：最小树形图的森林。因此我们不能确定根节点，因为根节点可以有多个，那怎么办呢？

3.设置一个超级点，作为虚拟的根节点。把这个超级点连向每一个题目中的点。然后再用此超级点去跑zhuliu算法，得到一个最小树形图（智能的超级点，好评）。可知把超级点去掉后，就是我们要求的最小树形图的森林（当然我们不需要这样做，这里只是方便理解）。

 

代码一：

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXM = 1e6+;
 const int MAXN = 1e3+;

 struct Edge
 {
     int u, v, w;
 }edge[MAXM];

 int x[MAXN], y[MAXN], z[MAXN];
 int pre[MAXN], id[MAXN], vis[MAXN], in[MAXN];

 int zhuliu(int root, int n, int m)
 {
     int res = ;
     while(true)
     {
         for(int i = ; i<n; i++)
             in[i] = INF;
         for(int i = ; i<m; i++)
         if(edge[i].u!=edge[i].v && edge[i].w<in[edge[i].v])
         {
             pre[edge[i].v] = edge[i].u;
             in[edge[i].v] = edge[i].w;
         }

         for(int i = ; i<n; i++)
             if(i!=root && in[i]==INF)
                 return -;

         int tn = ;
         memset(id, -, sizeof(id));
         memset(vis, -, sizeof(vis));
         in[root] = ;
         for(int i = ; i<n; i++)
         {
             res += in[i];
             int v = i;
             while(vis[v]!=i && id[v]==- && v!=root)
             {
                 vis[v] = i;
                 v = pre[v];
             }
             if(v!=root && id[v]==-)
             {
                 for(int u = pre[v]; u!=v; u = pre[u])
                     id[u] = tn;
                 id[v] = tn++;
             }
         }
         if(tn==) break;
         for(int i = ; i<n; i++)
             if(id[i]==-)
                 id[i] = tn++;

         for(int i = ;  i<m; )
         {
             int v = edge[i].v;
             edge[i].u = id[edge[i].u];
             edge[i].v = id[edge[i].v];
             if(edge[i].u!=edge[i].v)
                 edge[i++].w -= in[v];
             else
                 swap(edge[i], edge[--m]);
         }
         n = tn;
         root = id[root];
     }
     return res;
 }

 int main()
 {
     int n, m, X, Y, Z;
     while(scanf("%d%d%d%d", &n, &X, &Y, &Z)!=EOF)
     {
         if(!n && !X && !Y && !Z) break;

         for(int i = ; i<n; i++)
             scanf("%d%d%d", &x[i], &y[i], &z[i]);

         m = ;
         for(int i = ; i<n; i++)
         {
             int num, v;
             scanf("%d", &num);
             while(num--)
             {
                 scanf("%d", &v); v--;
                 edge[m].u = i;
                 edge[m].v = v;
                 edge[m].w = Y*(abs(x[i]-x[v])+abs(y[i]-y[v])+abs(z[i]-z[v]));
                 if(z[i]<z[v]) edge[m].w += Z;
                 m++;
             }
             //0~n-1为题目中的点， n为人工设置的超级点
             //将超级点连到每一个结点，并且设置相应的权值
             edge[m].u = n;
             edge[m].v = i;
             edge[m++].w = z[i]*X;
         }

         int ans = zhuliu(n, n+, m);
         printf("%d\n", ans);
     }
 }

代码二：

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXM = 1e6+;
 const int MAXN = 1e3+;

 struct Edge
 {
     int u, v, w;
 }edge[MAXM];

 int x[MAXN], y[MAXN], z[MAXN];
 int pre[MAXN], id[MAXN], vis[MAXN], in[MAXN];

 int zhuliu(int root, int n, int m)
 {
     int res = ;
     while(true)
     {
         for(int i = ; i<n; i++)
             in[i] = INF;
         for(int i = ; i<m; i++)
         if(edge[i].u!=edge[i].v && edge[i].w<in[edge[i].v])
         {
             pre[edge[i].v] = edge[i].u;
             in[edge[i].v] = edge[i].w;
         }

         for(int i = ; i<n; i++)
             if(i!=root && in[i]==INF)
                 return -;

         int tn = ;
         memset(id, -, sizeof(id));
         memset(vis, -, sizeof(vis));
         in[root] = ;
         for(int i = ; i<n; i++)
         {
             res += in[i];
             int v = i;
             while(vis[v]!=i && id[v]==- && v!=root)
             {
                 vis[v] = i;
                 v = pre[v];
             }
             if(v!=root && id[v]==-)
             {
                 for(int u = pre[v]; u!=v; u = pre[u])
                     id[u] = tn;
                 id[v] = tn++;
             }
         }
         if(tn==) break;
         for(int i = ; i<n; i++)
             if(id[i]==-)
                 id[i] = tn++;

         for(int i = ;  i<m; i++)
         {
             int v = edge[i].v;
             edge[i].u = id[edge[i].u];
             edge[i].v = id[edge[i].v];
             if(edge[i].u!=edge[i].v)
                 edge[i].w -= in[v];
         }
         n = tn;
         root = id[root];
     }
     return res;
 }

 int main()
 {
     int n, m, X, Y, Z;
     while(scanf("%d%d%d%d", &n, &X, &Y, &Z)!=EOF)
     {
         if(!n && !X && !Y && !Z) break;

         for(int i = ; i<n; i++)
             scanf("%d%d%d", &x[i], &y[i], &z[i]);

         m = ;
         for(int i = ; i<n; i++)
         {
             int num, v;
             scanf("%d", &num);
             while(num--)
             {
                 scanf("%d", &v); v--;
                 edge[m].u = i;
                 edge[m].v = v;
                 edge[m].w = Y*(abs(x[i]-x[v])+abs(y[i]-y[v])+abs(z[i]-z[v]));
                 if(z[i]<z[v]) edge[m].w += Z;
                 m++;
             }
             //0~n-1为题目中的点， n为人工设置的超级点
             //将超级点连到每一个结点，并且设置相应的权值
             edge[m].u = n;
             edge[m].v = i;
             edge[m++].w = z[i]*X;
         }

         int ans = zhuliu(n, n+, m);
         printf("%d\n", ans);
     }
 }