HDU-4009 Transfer Water (minimum tree shape)

Main topic: There are n points, each point has the corresponding three-dimensional coordinates (x, y, z)
Now there are two ways to get water at every point, or water.
1. Dig your own well, at a cost of x * Altitude Z
2. Laying pipelines for diversion.
A. If the altitude is less than the water diversion, the cost is between two Manhattan distances *y
B. If the altitude is greater than the drinking place, the cost is two Manhattan distance *y + Z

Problem-solving ideas: Set a virtual root, the virtual root leads to all points, the weight of the cost of digging well, and then according to the requirements of the edge, to find the smallest tree chart can be

#include <cstdio>#include <cstring>#define N 1010#define ABS (a) ((a) >0? ( A):(-(a)))structedge{intU, V, c;} E[n*n];structpoint{intX, Y, Z;} P[n];intn, x, y, z, tot;voidAddedge (intUintVintc) {e[tot].u = u; E[TOT].V = v; E[TOT++].C = C;}voidInit () {tot =0; for(inti =1; I <= N; i++) {scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].z); Addedge (0, I, p[i].z * x); }intK, T; for(inti =1; I <= N; i++) {scanf("%d", &k); while(k--) {scanf("%d", &t);if(t = = i)Continue;if(P[i].z >= p[t].z) {intdis =ABS(p[t].x-p[i].x) +ABS(P[T].Y-P[I].Y) +ABS(P[T].Z-P[I].Z);            Addedge (i, t, dis * y); }Else{intdis =ABS(p[t].x-p[i].x) +ABS(P[T].Y-P[I].Y) +ABS(P[T].Z-P[I].Z);            Addedge (i, t, dis * y + z); }        }    }}#define INF 0x3f3f3f3fintIn[n], Pre[n], vis[n], id[n];intDirected_mst (intRootintN) {intAns =0, u, V, TMP; while(1) { for(inti =0; I < n; i++) In[i] = INF; for(inti =0; i < tot;            i++) {u = e[i].u; v = e[i].v;if(U! = v && e[i].c < in[v])                {In[v] = e[i].c;            PRE[V] = u; }        }memset(Vis,-1,sizeof(VIS));memset(ID,-1,sizeof(ID)); In[root] =0;intsubnode =0; for(inti =0; I < n;            i++) {ans + = in[i]; TMP = i; while(vis[tmp]! = i && tmp! = root && id[tmp] = =-1) {Vis[tmp] = i;            TMP = pre[tmp]; }if(Id[tmp] = =-1&& tmp! = root) {u = pre[tmp]; while(U! = tmp)                    {Id[u] = subnode;                U = pre[u];            } id[tmp] = subnode++; }        }if(subnode = =0) Break; for(inti =0; I < n; i++)if(!                (~id[i])) Id[i] = subnode++; for(inti =0; i < tot;            i++) {tmp = E[I].V;            E[I].U = id[e[i].u]; E[I].V = id[e[i].v];if(e[i].u! = e[i].v)        E[I].C-= in[tmp];        } n = subnode;    root = Id[root]; }returnAns;}voidSolve () {n++;intAns = directed_mst (0, n);if(ans = =-1)printf("Poor xiaoa\n");Else        printf("%d\n", ans);}intMain () { while(scanf("%d%d%d%d", &n, &x, &y, &z)! = EOF && n + x + y + z) {init ();    Solve (); }return 0;}

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HDU-4009 Transfer Water (minimum tree shape)