"Tree dp" HDU 4303 Hourai Jeweled

Channel

Test instructions: Give a tree, the side has color, point the right value. A path that satisfies no two adjacent edges on a path is a good path, and the sum of the paths (the weights and the points on the path)

Ideas:

Benquan sort, starting at any point, deep search, after each subtree is searched, returns the number of pair< that can be extended to this point and the last edge is different from the edge color of the parent node to that point, all of which are the sum of the edge fractions >
After each deep search for a child node, the total score of the gorgeous edge added to this point is:
The sum of the number of edges returned by all the child nodes that were previously searched up * The score of the current child node +
The sum of the points returned by all the child nodes that were previously searched upward * The number of edges returned by the current child node +
The sum of the number of edges returned by all the child nodes that were previously searched * The number of edges returned by the current child node * Right of the current point

Code:

#include <cstdio>#include<cstring>#include<algorithm>using namespacestd;Const intMax_n =300007; typedefLong Longll;structEdge {intu, V, W, NXT; Edge () {} Edge (int_u,int_v,int_w,int_n) {u= _u; v = _v; W = _w; NXT =_n; }    BOOL operator< (ConstEdge &AMP;RHS)Const {        returnW <RHS.W; }};intN;intHead[max_n], ecnt;ll ans, val[max_n], Num[max_n], sum[max_n]; Edge T[max_n], g[max_n<<1];voidClear () {ecnt= ans =0; Memset (Head,-1,sizeofhead); memset (num,0,sizeofNUM); memset (SUM,0,sizeofsum);}voidAddintUintVintW) {g[ecnt]=Edge (U, V, W, Head[u]); Head[u]= ecnt++;}voidDfsintUintFaintCol) {ll tot=0, tot1 =0, Tot2 =0, k =0, K1 =0, K2 =0; intNowcol =0;  for(inti = Head[u]; ~i; i =g[i].nxt) {        intv = g[i].v, w =G[I].W; if(v = = FA)Continue;        DFS (V, u, W); if(w! = col) tot + = Sum[v], k + =Num[v]; if(W! =Nowcol) {TOT1+ = Tot2; K1 + =K2; Tot2= Sum[v], K2 =Num[v]; Nowcol=W; } ElseTot2 + = Sum[v], K2 + =Num[v]; Ans+ = sum[v] * (k1 +1) + tot1 * Num[v] + val[u] * (k1 +1) *Num[v]; } Num[u]= ++k, Sum[u] = val[u] * k +tot;} Template<classT>inlineBOOLRD (T &ret) {    CharCintSGN; if(c = GetChar (), c = = EOF)return false;  while(c! ='-'&& (C <'0'|| C >'9')) C =GetChar (); SGN= (c = ='-') ? -1:1; RET= (c = ='-') ?0: (C-'0');  while(c = GetChar (), C >='0'&& C <='9') ret = RET *Ten+ (C-'0'); RET*=SGN; return true;}intMain () { while(1= = scanf ("%d", &N)) { for(inti =1; I <= N; ++i) Rd (Val[i]);  for(inti =0; i +1< n; ++i) {rd (t[i].u), RD (T[I].V), RD (T[I].W); } sort (T, T+ N-1);        Clear ();  for(inti =0; i +1< n; ++i) {Add (t[i].u, T[I].V, T[I].W);        Add (T[I].V, t[i].u, T[I].W); } DFS (1, -1, -1); printf ("%i64d\n", ans); }    return 0;}

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Tree dp HDU 4303 Hourai Jeweled