HDU4009 Transfer water【最小樹形圖】【不定根】

標籤：

Transfer water

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 3943    Accepted Submission(s): 1415
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

題目大意：一個村莊被洪水摧毀了，整個村莊都要轉移。但是山上沒有泉水，每戶家庭只能

在自家挖一個水井或是修一個水渠從別的家庭引水。如果要修井，則修井費用和房子所在海

拔高度有關，每米X元。如果從別人的家裡引水，如果從高於自己家高度的人家裡引水，費

用為每米Y元。如果從低於自己家高度的人家裡引水，每條要多花費Z元。現在給你這個村莊

N個家庭房屋的座標(a,b,c)和三種花費X,Y,Z。接著給你各家之間能單向修建引水溝渠的限制。

問：能使全村莊的人喝上水的總修建費用最低為多少。若不能，則輸出"poor XiaoA"。

思路：其實就是給你一個有向圖，求有向圖的最小樹形圖是多少。但是根結點不確定。在這

裡可以假設一個根結點，也就是虛根，讓所有家庭都從虛根引水(其實就是每家都自己修井)，

所以問題肯定有解，最後就是朱劉演算法(模板)直接求最小樹形圖。

朱劉演算法參考我的另一篇博文：http://blog.csdn.net/lianai911/article/details/42242371

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>using namespace std;const int MAXN = 1010;const int MAXM = 1000010;struct Node{    int from;    int to;    int w;};Node Edges[MAXM];struct Node1{    int x;    int y;    int z;};Node1 Point[MAXN];int Dist(Node1 a,Node1 b){    return abs(a.x-b.x) + abs(a.y-b.y) + abs(a.z-b.z);}int pre[MAXN],vis[MAXN],flag[MAXN],In[MAXN],sum;int ZhuLiu(int root,int N,int M){    sum = 0;    while(true)    {        for(int i = 0; i < N; ++i)            In[i] = INT_MAX;        for(int i = 0; i < M; ++i)        {            int u = Edges[i].from;            int v = Edges[i].to;            if(Edges[i].w < In[v] && u != v)            {                pre[v] = u;                In[v] = Edges[i].w;            }        }        for(int i = 0; i < N; ++i)        {            if(i == root)                continue;            if(In[i] == INT_MAX)                return -1;        }        int CntNode = 0;        memset(flag,-1,sizeof(flag));        memset(vis,-1,sizeof(vis));        In[root] = 0;        for(int i = 0; i < N; ++i)        {            sum += In[i];            int v = i;            while(vis[v] != i && flag[v] == -1 && v != root)            {                vis[v] = i;                v = pre[v];            }            if(v != root && flag[v] == -1)            {                for(int u = pre[v]; u != v; u = pre[u])                    flag[u] = CntNode;                flag[v] = CntNode++;            }        }        if(CntNode == 0)            break;        for(int i = 0; i < N; ++i)        {            if(flag[i] == -1)                flag[i] = CntNode++;        }        for(int i = 0; i < M; ++i)        {            int v = Edges[i].to;            Edges[i].from = flag[Edges[i].from];            Edges[i].to = flag[Edges[i].to];            if(Edges[i].from != Edges[i].to)                Edges[i].w -= In[v];        }        N = CntNode;        root = flag[root];    }    return sum;}int main(){    int N,M,X,Y,Z,k,d;    while(~scanf("%d%d%d%d",&N,&X,&Y,&Z) && (N||X||Y||Z))    {        for(int i = 0; i < N; ++i)            scanf("%d%d%d",&Point[i].x,&Point[i].y,&Point[i].z);        M = 0;        for(int i = 0; i < N; ++i)        {            scanf("%d",&k);            while(k--)            {                scanf("%d",&d);                d--;                Edges[M].from = i;                Edges[M].to = d;                Edges[M].w = Dist(Point[i],Point[d])*Y;                if(Point[d].z > Point[i].z)                    Edges[M].w += Z;                M++;            }        }        for(int i = 0; i < N; ++i)        {            Edges[M].from = N;            Edges[M].to = i;            Edges[M++].w = Point[i].z*X;        }        printf("%d\n",ZhuLiu(N,N+1,M));    }    return 0;}

HDU4009 Transfer water【最小樹形圖】【不定根】