First, the idea
Enumerates all spanning tree Benquan and values, Benquan and value sum for each enumeration, modifies the Benquan of all edges (Es[i].cost-sum * 1.0/(N-1)) 2, i.e. the numerator of the variance formula, and then runs the minimum spanning tree algorithm while recording the original weights of the edges and, if the "minimum variance" is calculated The Benquan value of the spanning tree is sum, so use this "minimum variance" to update the answer.
Second, the complexity of analysis
Time complexity: O (N * W * M * LOGM). N * W is the time to enumerate the edge weights and values. The Benquan and values are minimum 0 and the maximum is (N-1) * W.
Third, PS
The problem is said to be Blue Bridge Cup official data. There are 5 examples (2, 3, 4, 5, 6), always WA. So, the right code can only get 50 points.
Iv. Source Code
#include <bits/stdc++.h>using namespacestd;intN, M;Const intMAXN = -, MAXM =1010;Const DoubleINF = (1LL << -) *1.0; typedefstructEDGE0 {intu, V, oldcost; DoubleNewcost; BOOL operator< (EDGE0 e)Const { returnNewcost <E.newcost; } voidAssgin (int_u,int_v,int_cost) {u=_u; V=_v; Oldcost=_cost; }} Edge; Edge edges[maxm];template<classT> InlinevoidRead (T &x) {intT; BOOLFlag =false; while((t = GetChar ())! ='-'&& (T <'0'|| T >'9')) ; if(T = ='-') flag =true, t =GetChar (); X= T-'0'; while(t = GetChar ()) >='0'&& T <='9') x = x *Ten+ T-'0'; if(flag) x =-x;}/** and check the set part*/intPAR[MAXN], RANK[MAXN];voidInit_ufind () { for(inti =0; i < MAXN; ++i) {Par[i]=i; Rank[i]=0; }}intUfind (intx) {returnx = = Par[x]? X:PAR[X] =Ufind (par[x]);}voidUniteintXinty) {x= Ufind (x), y =Ufind (y); if(x = = y)return; if(Rank[x] < rank[y]) par[x] =y; Else{Par[y]=x; if(Rank[x] = = Rank[y]) rank[x]++; }}BOOLSame (intXinty) {returnUfind (x) = =Ufind (y);}/** and check the set part*/DoubleKruscal (inttot) { DoubleAVG = tot *1.0/(N-1); for(inti =0; i < M; ++i) {edges[i].newcost= (Edges[i].oldcost *1.0-avg) * (Edges[i].oldcost *1.0-avg); } sort (edges, edges+M); Init_ufind (); Doubleres =0; intIres =0; for(inti =0; i < M; ++i) {Edge& e =Edges[i]; if(!Same (E.U, E.V)) {Unite (E.U, E.V); Res+=E.newcost; Ires+=E.oldcost; } } if(Ires = = tot)returnRES/(N-1); Else returnINF;}intMain () {#ifndef Online_judge freopen ("Input.txt","R", stdin);#endif intT =1, A, B, C; intCOSTS[MAXM]; while(SCANF ("%d%d", &n, &m),! (N = =0&& M = =0)) { for(inti =0; i < M; ++i) {read (a), read (b), read (c); EDGES[I].U= A, edges[i].v = b, edges[i].oldcost =C; Costs[i]=C; } sort (costs, costs+M); intMintot =0, Maxtot =0; for(inti =0; I < N-1; ++i) Mintot + =Costs[i]; for(inti = M-1; i > M-n;--i) maxtot + =Costs[i]; DoubleAns =INF; for(inttot = Mintot; Tot <= Maxtot; ++tot) {ans=min (ans, kruscal (tot)); } printf ("Case %d:%.2f\n", t++, ans); } return 0;}
Blue Bridge Cup--algorithm to improve the least squares of poor students into trees