HDU

HDU - 4009:http://acm.hdu.edu.cn/showproblem.php?pid=4009

题意：

　　　　有n户人家住在山上，现在每户人家（x，y，z）都要解决供水的问题，他可以自己挖井，也可以从特定的别人那里调水。问所有人家都接上水后最小的花费。

思路：

　　　　据说是一道最小生成树的模版题，我觉得在建图上还是比较难的。每个人家从别人那里调水的关系是明确的，直接建图就行。那自己挖井，我们可以建立一个虚拟的源点，向每个点连一条边，边的权值就是挖井所要的费用。建完图后，就可以跑一遍最小树形图了。显然答案是一定存在的；

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;       
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 10007;
const double esp = 1e-8;
const double PI=acos(-1.0);



template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/
            const int maxm = 1e6+9;
            const int maxn = 1e3+9;
            struct edge
            {
                int from,to,c;
            }e[maxm];
            int X,Y,Z,n,tot;
            struct node
            {
                int x,y,z;
            }p[maxn];
            int in[maxn],vis[maxn],pre[maxn],id[maxn];
            int rtt;
                int zhuliu(int root,int n,int m){
                    int res = 0;
                    while(true){
                        memset(in, inf, sizeof(in));
                        for(int i=1; i<=m ;i++){
                            if(e[i].from != e[i].to && e[i].c < in[e[i].to]){
                                pre[e[i].to] = e[i].from;
                                in[e[i].to] = e[i].c;
                            }
                        }
                        for(int i=1; i<=n; i++){
                            if(i!=root && in[i] == inf)
                                return -1;
                        }
                        int tn = 0,v;
                        memset(id,-1,sizeof(id));
                        memset(vis,-1,sizeof(vis));
                        in[root] = 0;
                        for(int i=1; i<=n; i++){
                            res += in[i];
                            v = i;
                            while(v!=root && id[v] == -1 && vis[v] != i){
                                vis[v] = i;
                                v = pre[v];
                            }
                            if(v!=root && id[v] == -1){
                                id[v] = ++tn;
                                for(int u = pre[v] ; u!=v; u = pre[u]){
                                    id[u] = tn;
                                }
                            }
                        }
                        if(tn == 0)break;
                        for(int i=1; i<=n; i++){
                            if(id[i] == -1)id[i] = ++tn;
                        }

                        for(int i=1; i<=m; i++){
                            int v = e[i].to;
                            e[i].to = id[e[i].to];
                            e[i].from = id[e[i].from];
                            if(e[i].to != e[i].from){
                                e[i].c -= in[v];
                            }
                        }
                        n = tn;root = id[root];
                    }
                    return res;
                }
                int get(int u,int v){
                    int ans = abs(p[u].x - p[v].x) +abs(p[u].y - p[v].y)+abs(p[u].z - p[v].z); 
                    ans = ans * Y;
                    if(p[v].z > p[u].z)ans += Z;
                    return ans;
                }
int main(){
                while(~scanf("%d%d%d%d", &n, &X,&Y,&Z) && n+X+Y+Z){
                    tot = 0;
                    for(int i=1; i<=n; i++){
                        scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].z);
                    }
                    for(int i=1; i<=n; i++){
                        int x,k;scanf("%d", &k);
                        for(int j=1; j<=k; j++)
                        {
                            scanf("%d", &x);
                            e[++tot].from = i;
                            e[tot].to = x;
                            e[tot].c = get(i,x);
                        }
                    }

                    for(int i=1; i<=n; i++){
                        e[++tot].from = 0;
                        e[tot].to = i;
                        e[tot].c = X * p[i].z;
                    }   

                    int ans = zhuliu(0,n,tot);
                    printf("%d
", ans);
                }
                return 0;
}