樹上倍增：

　　沒講的，直接上題

【SCOI2016】幸運數字（題目）：

　　倍增時合併一下線性基即可

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<iostream>
  4 #include<algorithm>
  5 using namespace std;
  6 
  7 typedef long long LL;
  8 
  9 const int N = 20010;
 10 
 11 int n, q;
 12 LL G[N], dep[N], ans[N];
 13
 
 LL fa[N][21], p[N][21][62];
 14 int h[N], num[N << 1], nex[N << 1], dqx;
 15 
 16 inline void add(int a, int b)
 17 {
 18     num[dqx] = b;
 19     nex[dqx] = h[a];
 20     h[a] = dqx++;
 21 }
 22 
 23 inline void get_p(LL *a, LL val)
 24 {
 25     for (int i = 61; i >= 0; i--)
 26     {
 
 27         if ((val >> i) & 1)
 28         {
 29             if (!a[i])
 30             {
 31                 a[i] = val;
 32                 break;
 33             }
 34             val ^= a[i];
 35         }
 36     }
 37 }
 38 
 39 void dfs(int u, int f)
 40 {
 41     fa[u][0] = f;
 42     dep[u] = dep[f] + 1
 
;
 43     for (int i = h[u]; ~i; i = nex[i])
 44     {
 45         int j = num[i];
 46         if (j == f) continue;
 47         dfs(j, u);
 48     }
 49 }
 50 
 51 void merge(LL* a, LL* b)
 52 {
 53     for (int i = 61; i >= 0; i--)
 54     {
 55         if (b[i]) get_p(a, b[i]);
 56     }
 57 }
 58 
 59 void get_lca()
 60 {
 61     for (int j = 1; j < 20; j++)
 62     {
 63         for (int i = 1; i <= n; i++)
 64         {
 65             fa[i][j] = fa[fa[i][j - 1]][j - 1];
 66             memcpy(p[i][j], p[i][j - 1], sizeof(p[i][j - 1]));
 67             merge(p[i][j], p[fa[i][j - 1]][j - 1]);
 68         }
 69     }
 70 }
 71 
 72 inline void LCA(int u, int v)
 73 {
 74     if (dep[u] < dep[v]) swap(u, v);
 75 
 76     for (int i = 20; i >= 0; i--)
 77     {
 78         if (dep[fa[u][i]] >= dep[v])
 79         {
 80             merge(ans, p[u][i]);
 81             u = fa[u][i];
 82         }
 83     }
 84         
 85     if (u == v)
 86     {
 87         merge(ans, p[u][0]);
 88         return;
 89     }
 90 
 91     for (int i = 20; i >= 0; i--)
 92     {
 93         if (fa[u][i] != fa[v][i])
 94         {
 95             merge(ans, p[u][i]), merge(ans, p[v][i]);
 96             u = fa[u][i], v = fa[v][i];
 97         }
 98     }
 99         
100     merge(ans, p[u][0]);
101     merge(ans, p[v][0]);
102     merge(ans, p[fa[u][0]][0]);
103 }
104 
105 int main()
106 {
107     scanf("%d%d", &n, &q);
108     for (int i = 1; i <= n; i++)
109     {
110         scanf("%lld", &G[i]);
111         get_p(p[i][0], G[i]);
112     }
113 
114     memset(h, -1, sizeof(h));
115     for (int i = 1; i <= n - 1; i++)
116     {
117         int a, b;
118         scanf("%d%d", &a, &b);
119         add(a, b), add(b, a);
120     }
121 
122     dfs(1, 0);
123     get_lca();
124 
125     for (int i = 1; i <= q; i++)
126     {
127         memset(ans, 0, sizeof(ans));
128 
129         int a, b;
130         scanf("%d%d", &a, &b);
131         LCA(a, b);
132         LL res = 0;
133 
134         for (int j = 61; j >= 0; j--) res = max(res, res ^ (LL)ans[j]);
135         printf("%lld\n", res);
136     }
137 }