HDU 4009 不定根最小树形图

讲一下建图过程，首先建立一个超级源点S，对于这个源点，向每个HOUSE连一条有向边，权值为该HOUSE建立WELL的费用，即高度*X。

然后每个可以连边的WELL之间，费用为曼哈顿距离*Y，然后考虑两边的高度，如果需要连接PUMB，则在该费用上+Z。

这样建图之后，以S为根，跑一遍最小树形图算法即可。

CODE：

#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <iostream>#include <algorithm>#define Max 2505#define FI first#define SE second#define ll long long#define PI acos(-1.0)#define inf 0x3fffffff#define LL(x) ( x << 1 )#define bug puts("here")#define PII pair<int,int>#define RR(x) ( x << 1 | 1 )#define mp(a,b) make_pair(a,b)#define mem(a,b) memset(a,b,sizeof(a))#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )using namespace std;inline void RD(int &ret) {    char c;    int flag = 1 ;    do {        c = getchar();        if(c == '-')flag = -1 ;    } while(c < '0' || c > '9') ;    ret = c - '0';    while((c=getchar()) >= '0' && c <= '9')        ret = ret * 10 + ( c - '0' );    ret *= flag ;}inline void OT(int a) {    if(a >= 10)OT(a / 10) ;    putchar(a % 10 + '0') ;}/*********************************************/#define N 1005int n , X , Y , Z ;struct C{    int x , y , z ;}city[N] ;int num = 0 ;int S ;void init(){    num = S = 0 ;}struct kdq{    int s ,e ,l ;}E[N * N] ;int get_Mandis(int i ,int j){    return abs(city[i].x - city[j].x) + abs(city[i].y - city[j].y) + abs(city[i].z - city[j].z) ;}void add(int s ,int e ,int l){    E[num].s = s ;    E[num].e = e ;    E[num].l = l ;    num ++ ;}int pre[N] , vis[N] , id[N] , in[N] ;int Directed_MST(int root ,int NV ,int NE){    int ret = 0 ;    while(1){        for (int i = 0 ; i < NV ; i ++ )in[i] = inf ;        for (int i = 0 ; i < NE ; i ++ ){            int s = E[i].s ;            int e = E[i].e ;            if(in[e] > E[i].l && s != e){                in[e] = E[i].l ;                pre[e] = s ;            }        }        for (int i = 0 ; i < NV ; i ++ ){            if(i == root)continue ;            if(in[i] == inf)return -1 ;        }        int cntnode = 0 ;        mem(vis , -1) ;        mem(id , -1) ;        in[root] = 0 ;        for (int i = 0 ; i < NV ; i ++ ){            ret += in[i] ;            int v = i ;            while(vis[v] != i && id[v] == -1 && v != root){                vis[v] = i ;                v = pre[v] ;            }            if(v != root && id[v] == -1){                for (int u = pre[v] ; u != v ; u = pre[u]){                    id[u] = cntnode ;                }                id[v] = cntnode ++ ;            }        }        if(cntnode == 0)break ;        for (int i = 0 ; i < NV ; i ++ )if(id[i] == -1)id[i] = cntnode ++ ;        for (int i = 0 ; i < NE ; i ++ ){            int s = E[i].s ;            int e = E[i].e ;            E[i].s = id[s] ;            E[i].e = id[e] ;            if(id[s] != id[e])E[i].l -= in[e] ;        }        NV = cntnode ;        root = id[root] ;    }    return ret ;}int main() {    int a , k ;    while(scanf("%d%d%d%d",&n,&X,&Y,&Z) , ( n + X + Y + Z)){        init() ;        for (int i = 1 ; i <= n ;i ++ ){            RD(city[i].x) ;RD(city[i].y) ;RD(city[i].z) ;        }        for (int i = 1 ; i <= n ; i ++ ){            RD(k) ;            while(k -- ){                RD(a) ;                if(i == a)continue ;//自环                int dis = get_Mandis(i , a) * Y ;                if(city[i].z < city[a].z)dis += Z ;                add(i , a , dis) ;            }        }        for (int i = 1 ; i <= n ; i ++ ){            add(S , i , city[i].z * X) ;        }        int ans = Directed_MST(0 , n + 1 , num) ;        if(ans == -1)puts("poor XiaoA") ;        else OT(ans) ;        puts("") ;    }    return 0 ;}