标签：des style blog http color io os ar strong
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 3875 | Accepted: 850 |
Description
In an edge-weighted tree, the xor-length of a path p is defined as the xor sum of the weights of edges on p:
⊕ is the xor operator.
We say a path the xor-longest path if it has the largest xor-length. Given an edge-weighted tree with n nodes, can you find the xor-longest path?
Input
The input contains several test cases. The first line of each test case contains an integer n(1<=n<=100000), The following n-1 lines each contains three integers u(0 <= u < n),v(0 <= v < n),w(0 <= w < 2^31), which means there is an edge between node u and v of length w.
Output
Sample Input
4 0 1 3 1 2 4 1 3 6
Sample Output
7
Hint
The xor-longest path is 0->1->2, which has length 7 (=3 ⊕ 4)
這道題由於沒看到多組數據，WA了很久，題目比較簡單，但是運用了a xor b= a xor c xor b xor c，這應該是解異或題目常用的技巧。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<ctime> #include<cmath> #include<algorithm> #include<set> #include<map> #include<vector> #include<string> #include<queue> #include<stack> using namespace std; #ifdef WIN32 #define LL "%I64d" #else #define LL "%lld" #endif #define MAXN 110000 #define MAXV MAXN #define MAXE MAXV*2 #define MAXT MAXN*200 //AC typedef long long qword; int n,m; struct trie { int ch[2]; }tree[MAXT]; int toptr=1; inline void new_node(int &now) { now=++toptr; tree[now].ch[0]=tree[now].ch[1]=0; } int dbg=0; void build_trie(qword x) { dbg++; int now=1; int i; for (i=32;i>=0;i--) { if (!tree[now].ch[(x&(1ll<<i))!=0]) { new_node(tree[now].ch[(x&(1ll<<i))!=0]); } now=tree[now].ch[(x&(1ll<<i))!=0]; } if (dbg==175672) dbg--; } struct Edge { int np; qword val; Edge *next; }E[MAXE],*V[MAXV]; int tope=-1; void addedge(int x,int y,qword z) { E[++tope].np=y; E[tope].val=z; E[tope].next=V[x]; V[x]=&E[tope]; } int fa[MAXN],depth[MAXN],val[MAXN]; void dfs(int now,int v) { Edge *ne; build_trie(v); val[now]=v; for (ne=V[now];ne;ne=ne->next) { if (ne->np==fa[now])continue; fa[ne->np]=now; dfs(ne->np,v^ne->val); } } qword find(qword x) { int now=1; int ret=0; int i; for (i=32;i>=0;i--) { if (tree[now].ch[(x&(1ll<<i))==0]) { now=tree[now].ch[(x&(1ll<<i))==0]; ret+=1ll<<i; }else { now=tree[now].ch[(x&(1ll<<i))!=0]; } } if (!now)throw 1; return ret; } int main() { freopen("input.txt","r",stdin); //freopen("output.txt","w",stdout); int i,j,k; qword x,y,z; while (~scanf("%d",&n)) { memset(V,0,sizeof(V)); memset(fa,-1,sizeof(fa)); toptr=1; tope=-1; tree[1].ch[0]=tree[1].ch[1]=0; for (i=1;i<n;i++) { scanf(LL LL LL,&x,&y,&z); addedge(x,y,z); addedge(y,x,z); } fa[0]=0; dfs(0,0); qword ans=0; for (i=0;i<n;i++) { ans=max(ans,find(val[i])); } printf(LL "\n",ans); } return 0; }
标签：des style blog http color io os ar strong
原文地址：http://www.cnblogs.com/mhy12345/p/3970458.html
