字典树;
这里字典树用于存储dis这些数。。。然后对于某个数x来说，要使得它变大，就要找到一个与它能“合适”与它异或的数。
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1 /* 2 字典树+异或 3 a^b = (a^c)^(b^c) 4 */ 5 #include<stdio.h> 6 #include<string.h> 7 #include<stdlib.h> 8 const int maxn = 100005; 9 const int maxm = 31; 10 int tree[ maxn*maxm ][ 2 ]; 11 int dis[ maxn ],vis[ maxn ]; 12 struct node{ 13 int u,next,val; 14 }; 15 node edge[ maxn<<2 ]; 16 int cnt,head[ maxn ]; 17 int tree_index; 18 19 void init(){ 20 memset( dis,0,sizeof( dis )); 21 memset( vis,0,sizeof( vis )); 22 cnt = 0; 23 memset( head,-1,sizeof( head ) ); 24 memset( tree,-1,sizeof( tree ) ); 25 } 26 27 void addedge( int a,int b,int c ){ 28 edge[ cnt ].u = b; 29 edge[ cnt ].val = c; 30 edge[ cnt ].next = head[ a ]; 31 head[ a ] = cnt++; 32 } 33 34 void dfs( int now ){ 35 for( int i=head[ now ];i!=-1;i=edge[ i ].next ){ 36 int u = edge[ i ].u; 37 if( vis[ u ] == 0 ){ 38 vis[ u ] = 1; 39 dis[ u ] = dis[ now ]^edge[ i ].val; 40 dfs( u ); 41 } 42 } 43 return ; 44 }//求出其他点到根的异或值 45 46 int find_the_max( int aim ){ 47 int res = 0; 48 int id = 0; 49 for( int i=30;i>=0;i-- ){ 50 int one_bit = (aim>>i)&1; 51 if( one_bit==0 ){ 52 if( tree[ id ][ 1 ]!=-1 ){ 53 res += 1<<i; 54 id = tree[ id ][ 1 ]; 55 } 56 else if( tree[ id ][ 0 ]!=-1 ) id = tree[ id ][ 0 ]; 57 else break; 58 } 59 else if( one_bit==1 ){ 60 if( tree[ id ][ 0 ]!=-1 ){ 61 res += 1<<i; 62 id = tree[ id ][ 0 ]; 63 } 64 else if( tree[ id ][ 1 ]!=-1 ) id = tree[ id ][ 1 ]; 65 else break; 66 } 67 } 68 return res; 69 }/*贪心，对于当前一个数x，异或某个数y之后变大，必须求一个数从高位到低位和x不同的。。*/ 70 71 void rebuild_tree( int aim ){ 72 int id = 0; 73 for( int i=30;i>=0;i-- ){ 74 int one_bit = (aim>>i)&1; 75 if( tree[ id ][ one_bit ]==-1 ){ 76 tree[ id ][ one_bit ] = tree_index++; 77 } 78 id = tree[ id ][ one_bit ]; 79 } 80 return ; 81 }//把dis[]变为2进制，存入字典数（从当前的tree_index开始存储） 82 83 void solve( int n ){ 84 int ans = 0; 85 tree_index = 1; 86 for( int i=0;i<n;i++ ){ 87 int tmp_ans = find_the_max( dis[ i ] ); 88 if( tmp_ans>ans ) ans = tmp_ans; 89 rebuild_tree( dis[i] ); 90 } 91 printf("%d\n",ans); 92 return ; 93 } 94 95 int main(){ 96 int n; 97 while( scanf("%d",&n)!=EOF ){ 98 int a,b,c; 99 init(); 100 for( int i=0;i<n-1;i++ ){ 101 scanf("%d%d%d",&a,&b,&c); 102 addedge( a,b,c ); 103 addedge( b,a,c ); 104 } 105 106 vis[ 0 ] = 1; 107 dfs( 0 ); 108 //for( int i=0;i<n;i++ ) 109 // printf("dis[%d]=%d\n",i,dis[i]); 110 solve( n ); 111 112 } 113 return 0; 114 }