POJ

POJ - 3764
题意： 求树上两个结点间路径的XOR值的最大值。
思路： 可以知道任意两个结点，设从根到这个结点的dis[u] = XOR路径和，那么u->v = dis[u]XORdis[v];
因为他们的dis[lca(u,v)]会因为被XOR两次而不会对结果造成影响，所以这就成了求dis[i]中两个值的XOR最大值，可以用01字典树维护，每对一个dis[i]值，先查询再插入。

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 1e5 + 10;struct node{    int to,w,next;    node(){}    node(int a,int b,int c){to = a; w = b; next = c;}}edge[maxn << 1];int h[maxn];int ch[maxn*32][2], val[maxn*32], dis[maxn];int cnum,num,n;void init(){    for(int i = 0; i < n; i++) h[i] = -1;    memset(ch[0], 0 , sizeof(ch[0]));    cnum = num = 0;    dis[0] = 0;}void add(int u,int v,int w){    edge[num] = node(v,w,h[u]);h[u] = num++;}void dfs(int u,int fa){    for(int i = h[u];  ~i ; i = edge[i].next)    {        int v = edge[i].to,w = edge[i].w;        if(v == fa) continue;        dis[v] = dis[u] ^ w;        dfs(v,u);    }}void Insert(int x){    int cur = 0;    for(int i = 31; i >= 0; i--)    {        int idx = (x >> i)&1;        if(!ch[cur][idx])        {            memset(ch[cnum],0,sizeof(ch[cnum]));            ch[cur][idx] = cnum;            val[cnum++] = 0;        }        cur = ch[cur][idx];    }    val[cur] = x;}int query(int x){    int cur = 0;    for(int i = 31; i >= 0; i--)    {        int idx = (x >> i)&1;        if(ch[cur][idx^1]) cur = ch[cur][idx^1];        else cur = ch[cur][idx];    }    return val[cur]^x;}int main(){    int a,b,c;    while(~scanf("%d",&n))    {        init();        for(int i = 0; i < n-1; i++)        {            scanf("%d%d%d",&a,&b,&c);            add(a,b,c); add(b,a,c);        }        dfs(0,-1);        int ans = 0;        Insert(0);        for(int i = 0; i < n; i++)        {            ans = max(ans,query(dis[i]));            Insert(dis[i]);        }        printf("%d\n",ans);    }    return 0;}