HDU4009 Transfer water —— 最小树形图 + 超级点

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标签：mit cstring log min += field ring water sub

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

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

代码一：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <algorithm>
  4 #include <cstring>
  5 #include <cmath>
  6 using namespace std;
  7 typedef long long LL;
  8 const double EPS = 1e-6;
  9 const int INF = 2e9;
 10 const LL LNF = 9e18;
 11 const int MOD = 1e9+7;
 12 const int MAXM = 1e6+10;
 13 const int MAXN = 1e3+10;
 14 
 15 struct Edge
 16 {
 17     int u, v, w;
 18 }edge[MAXM];
 19 
 20 int x[MAXN], y[MAXN], z[MAXN];
 21 int pre[MAXN], id[MAXN], vis[MAXN], in[MAXN];
 22 
 23 LL zhuliu(int root, int n, int m)
 24 {
 25     LL res = 0;
 26     while(true)
 27     {
 28         for(int i = 0; i<n; i++)
 29             in[i] = INF;
 30         for(int i = 0; i<m; i++)
 31         if(edge[i].u!=edge[i].v && edge[i].w<in[edge[i].v])
 32         {
 33             pre[edge[i].v] = edge[i].u;
 34             in[edge[i].v] = edge[i].w;
 35         }
 36 
 37         for(int i = 0; i<n; i++)
 38             if(i!=root && in[i]==INF)
 39                 return -1;
 40 
 41         int tn = 0;
 42         memset(id, -1, sizeof(id));
 43         memset(vis, -1, sizeof(vis));
 44         in[root] = 0;
 45         for(int i = 0; i<n; i++)
 46         {
 47             res += in[i];
 48             int v = i;
 49             while(vis[v]!=i && id[v]==-1 && v!=root)
 50             {
 51                 vis[v] = i;
 52                 v = pre[v];
 53             }
 54             if(v!=root && id[v]==-1)
 55             {
 56                 for(int u = pre[v]; u!=v; u = pre[u])
 57                     id[u] = tn;
 58                 id[v] = tn++;
 59             }
 60         }
 61         if(tn==0) break;
 62         for(int i = 0; i<n; i++)
 63             if(id[i]==-1)
 64                 id[i] = tn++;
 65 
 66         for(int i = 0;  i<m; )
 67         {
 68             int v = edge[i].v;
 69             edge[i].u = id[edge[i].u];
 70             edge[i].v = id[edge[i].v];
 71             if(edge[i].u!=edge[i].v)
 72                 edge[i++].w -= in[v];
 73             else
 74                 swap(edge[i], edge[--m]);
 75         }
 76         n = tn;
 77         root = id[root];
 78     }
 79     return res;
 80 }
 81 
 82 int main()
 83 {
 84     int n, m, X, Y, Z;
 85     while(scanf("%d%d%d%d", &n, &X, &Y, &Z)!=EOF)
 86     {
 87         if(!n && !X && !Y && !Z) break;
 88 
 89         for(int i = 0; i<n; i++)
 90             scanf("%d%d%d", &x[i], &y[i], &z[i]);
 91 
 92         m = 0;
 93         for(int i = 0; i<n; i++)
 94         {
 95             int num, v;
 96             scanf("%d", &num);
 97             while(num--)
 98             {
 99                 scanf("%d", &v); v--;
100                 edge[m].u = i;
101                 edge[m].v = v;
102                 edge[m].w = Y*(abs(x[i]-x[v])+abs(y[i]-y[v])+abs(z[i]-z[v]));
103                 if(z[i]<z[v]) edge[m].w += Z;
104                 m++;
105             }
106 
107             edge[m].u = n;
108             edge[m].v = i;
109             edge[m++].w = z[i]*X;
110         }
111 
112         LL ans = zhuliu(n, n+1, m);
113         printf("%I64d\n", ans);
114     }
115 }

View Code

代码二：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <algorithm>
  4 #include <cstring>
  5 #include <cmath>
  6 using namespace std;
  7 typedef long long LL;
  8 const double EPS = 1e-6;
  9 const int INF = 2e9;
 10 const LL LNF = 9e18;
 11 const int MOD = 1e9+7;
 12 const int MAXM = 1e6+10;
 13 const int MAXN = 1e3+10;
 14 
 15 struct Edge
 16 {
 17     int u, v, w;
 18 }edge[MAXM];
 19 
 20 int x[MAXN], y[MAXN], z[MAXN];
 21 int pre[MAXN], id[MAXN], vis[MAXN], in[MAXN];
 22 
 23 LL zhuliu(int root, int n, int m)
 24 {
 25     LL res = 0;
 26     while(true)
 27     {
 28         for(int i = 0; i<n; i++)
 29             in[i] = INF;
 30         for(int i = 0; i<m; i++)
 31         if(edge[i].u!=edge[i].v && edge[i].w<in[edge[i].v])
 32         {
 33             pre[edge[i].v] = edge[i].u;
 34             in[edge[i].v] = edge[i].w;
 35         }
 36 
 37         for(int i = 0; i<n; i++)
 38             if(i!=root && in[i]==INF)
 39                 return -1;
 40 
 41         int tn = 0;
 42         memset(id, -1, sizeof(id));
 43         memset(vis, -1, sizeof(vis));
 44         in[root] = 0;
 45         for(int i = 0; i<n; i++)
 46         {
 47             res += in[i];
 48             int v = i;
 49             while(vis[v]!=i && id[v]==-1 && v!=root)
 50             {
 51                 vis[v] = i;
 52                 v = pre[v];
 53             }
 54             if(v!=root && id[v]==-1)
 55             {
 56                 for(int u = pre[v]; u!=v; u = pre[u])
 57                     id[u] = tn;
 58                 id[v] = tn++;
 59             }
 60         }
 61         if(tn==0) break;
 62         for(int i = 0; i<n; i++)
 63             if(id[i]==-1)
 64                 id[i] = tn++;
 65 
 66         for(int i = 0;  i<m; i++)
 67         {
 68             int v = edge[i].v;
 69             edge[i].u = id[edge[i].u];
 70             edge[i].v = id[edge[i].v];
 71             if(edge[i].u!=edge[i].v)
 72                 edge[i].w -= in[v];
 73         }
 74         n = tn;
 75         root = id[root];
 76     }
 77     return res;
 78 }
 79 
 80 int main()
 81 {
 82     int n, m, X, Y, Z;
 83     while(scanf("%d%d%d%d", &n, &X, &Y, &Z)!=EOF)
 84     {
 85         if(!n && !X && !Y && !Z) break;
 86 
 87         for(int i = 0; i<n; i++)
 88             scanf("%d%d%d", &x[i], &y[i], &z[i]);
 89 
 90         m = 0;
 91         for(int i = 0; i<n; i++)
 92         {
 93             int num, v;
 94             scanf("%d", &num);
 95             while(num--)
 96             {
 97                 scanf("%d", &v); v--;
 98                 edge[m].u = i;
 99                 edge[m].v = v;
100                 edge[m].w = Y*(abs(x[i]-x[v])+abs(y[i]-y[v])+abs(z[i]-z[v]));
101                 if(z[i]<z[v]) edge[m].w += Z;
102                 m++;
103             }
104 
105             edge[m].u = n;
106             edge[m].v = i;
107             edge[m++].w = z[i]*X;
108         }
109 
110         LL ans = zhuliu(n, n+1, m);
111         printf("%I64d\n", ans);
112     }
113 }

View Code

HDU4009 Transfer water —— 最小树形图 + 超级点

标签：mit cstring log min += field ring water sub

原文地址：http://www.cnblogs.com/DOLFAMINGO/p/7768030.html

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