HDU - 4009 - Transfer water 朱刘算法 +建立虚拟节点
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 << "\n"; #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 '\n' #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\n", ans); } return 0; }
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