hdu 4009（最小有向生成树）

最新推荐文章于 2022-01-18 17:36:05 发布

Wiking__acm

最新推荐文章于 2022-01-18 17:36:05 发布

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Transfer water

Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 25 Accepted Submission(s): 2

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.

SampleInput

SampleOutput

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|. 
   
   
   
   
    
    加一个超级源点0，表示井，这样从井连向其他所有点的长度为那些点建井的代价。剩下的就是套模版了。
   
   
   
   
    
    #include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
using namespace std;
const int maxn = 1000 + 5;
const int INF = 1000000000;
typedef long long LL;

// 固定根的最小树型图，邻接矩阵写法，时间复杂度O（n^3）
struct MDST {
  int n;
  int w[maxn][maxn]; // 边权
  int vis[maxn];     // 访问标记，仅用来判断无解
  int ans;           // 计算答案
  int removed[maxn]; // 每个点是否被删除
  int cid[maxn];     // 所在圈编号
  int pre[maxn];     // 最小入边的起点
  int iw[maxn];      // 最小入边的权值
  int max_cid;       // 最大圈编号

  void init(int n) {
    this->n = n;
    for(int i = 0; i < n; i++)
      for(int j = 0; j < n; j++) w[i][j] = INF;
  }

  void AddEdge(int u, int v, int cost) {
    w[u][v] = min(w[u][v], cost); // 重边取权最小的
  }

  // 从s出发能到达多少个结点
  int dfs(int s) {
    vis[s] = 1;
    int ans = 1;
    for(int i = 0; i < n; i++)
      if(!vis[i] && w[s][i] < INF) ans += dfs(i);
    return ans;
  }

  // 从u出发沿着pre指针找圈
  bool cycle(int u) {
    max_cid++;
    int v = u;
    while(cid[v] != max_cid) { cid[v] = max_cid; v = pre[v]; }
    return v == u;
  }

  // 计算u的最小入弧，入弧起点不得在圈c中
  void update(int u) {
    iw[u] = INF;
    for(int i = 0; i < n; i++)
      if(!removed[i] && w[i][u] < iw[u]) {
        iw[u] = w[i][u];
        pre[u] = i;
      }
  }

  // 根结点为s，如果失败则返回false
  bool solve(int s) {
    memset(vis, 0, sizeof(vis));
    if(dfs(s) != n) return false;

    memset(removed, 0, sizeof(removed));
    memset(cid, 0, sizeof(cid));
    for(int u = 0; u < n; u++) update(u);
    pre[s] = s; iw[s] = 0; // 根结点特殊处理
    ans = max_cid = 0;
    for(;;) {
      bool have_cycle = false;
      for(int u = 0; u < n; u++) if(u != s && !removed[u] && cycle(u)){
        have_cycle = true;
        // 以下代码缩圈，圈上除了u之外的结点均删除
        int v = u;
        do {
          if(v != u) removed[v] = 1;
          ans += iw[v];
          // 对于圈外点i，把边i->v改成i->u（并调整权值）；v->i改为u->i
          // 注意圈上可能还有一个v'使得i->v'或者v'->i存在，因此只保留权值最小的i->u和u->i
          for(int i = 0; i < n; i++) if(cid[i] != cid[u] && !removed[i]) {
            if(w[i][v] < INF) w[i][u] = min(w[i][u], w[i][v]-iw[v]);
            w[u][i] = min(w[u][i], w[v][i]);
            if(pre[i] == v) pre[i] = u;
          }
          v = pre[v];
        } while(v != u);
        update(u);
        break;
      }
      if(!have_cycle) break;
    }
    for(int i = 0; i < n; i++)
      if(!removed[i]) ans += iw[i];
    return true;
  }
};

MDST solver;

struct Point{
    int x,y,h;
}p[maxn];

int abs(int x){return x<0?-x:x;}

int main(){
    int n;
    int X,Y,Z;
    while(scanf("%d%d%d%d",&n,&X,&Y,&Z)){
        if(n == 0 && X == 0 && Y == 0 && Z == 0) break;
        solver.init(n+1);
        for(int i = 1;i <= n;i++){
            scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].h);
            solver.AddEdge(0,i,p[i].h*X);
        }

        for(int i = 1;i <= n;i++){
            int num;scanf("%d",&num);
            while(num--){
                int to;scanf("%d",&to);
                if(to == i) continue;
                int tem = (abs(p[i].x-p[to].x)+abs(p[i].y-p[to].y)+abs(p[i].h-p[to].h)) * Y;
                if(p[i].h < p[to].h) solver.AddEdge(i,to,tem+Z);
                else solver.AddEdge(i,to,tem);
            }
        }

        if(solver.solve(0) == false) printf("poor XiaoA\n");
        else printf("%d\n",solver.ans);
    }
    return 0;
}