ACM Learning Journey-poj 3764 the Xor-longest Path (xor && dictionary tree && Greedy)

Title Link:http://poj.org/problem?id=3764

The main idea is to seek a path in the tree, making xor and maximum.

Because it is on a tree, there should be a unique path between the two nodes.

and xor (U, v) = XOR (0, U) ^xor (0, v).

So if you preprocess the XOR path of the 0 node to all nodes , the problem is converted to the number of n to take out two numbers, making the XOR Maximum.

This has previously dealt with a similar problem with a dictionary tree.

Code:

#include <iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<algorithm>#include<Set>#include<map>#include<queue>#include<string>#defineLL Long Longusing namespacestd;Const intMAXN =100005;//the maximum value of the two-energy XOR is taken out of the n number//Trie Tree Dictionary//Len*n, len number of binary maximum lengthsConst intLen = to;//Len indicates the binary maximum length of the numberstructtrie{intnext[2];} TREE[MAXN*Len];intCnttree;voidInittree () {Cnttree=0; memset (Tree,-1,sizeof(tree));voidAddintx) {    intnow =0; BOOLK;  for(inti = len; I >=0; i--) {k= x& (1<<i); if(Tree[now].next[k] = =-1) Tree[now].next[k]= ++Cnttree; now=Tree[now].next[k]; }}//returns the number of the maximum number of possible and X compositions in the current numberintQueryintx) {    intv =0, now =0; BOOLK;  for(inti = len; I >=0; i--) {k= x& (1<<i); if(Tree[now].next[!k]! =-1) K= !K; V= v| (k<<i); now=Tree[now].next[k]; }    returnv;}//chain-forward starstructedge{intto, next; intVal;} EDGE[MAXN*2];intHEAD[MAXN], CNT;voidAddedge (intUintVintW) {edge[cnt].to=v; Edge[cnt].next=Head[u]; Edge[cnt].val=W; Head[u]=CNT; CNT++;}voidInitedge () {memset (head,-1,sizeof(head)); CNT=0;}intN, P[MAXN];voidDfsintNow ) {    intto ;  for(inti = Head[now]; I! =-1; i =Edge[i].next) { to=edge[i].to; if(P[to] = =-1) {P[to]= p[now]^Edge[i].val;        DFS (to); }    }}voidinput () {Initedge (); Memset (P,-1,sizeof(p)); intu, V, W;  for(inti =1; I < n; ++i) {scanf ("%d%d%d", &u, &v, &W);        Addedge (U, V, W);    Addedge (V, U, W); } p[0] =0; DFS (0);}voidWork () {intAns =0;    Inittree ();  for(inti =0; I < n; ++i) {Add (P[i]); Ans= Max (ans, p[i]^query (P[i])); } printf ("%d\n", ans);}intMain () {//freopen ("test.in", "R", stdin);     while(SCANF ("%d", &n)! =EOF)        {input ();    Work (); }    return 0;}

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ACM Learning Journey-poj 3764 the Xor-longest Path (xor && dictionary tree && Greedy)