Solution
在步伐$pace$比较小的时候， 我们发现用前缀和直接维护会很快
而在$pace$比较大的时候， 则暴力往上跳会最优
设$blo= sqrt{N}$
若$pace<=blo$， 则利用前缀和更新,
预处理复杂度$O(N sqrt{N})$， 查询复杂度$O(1)$
若$pace>blo$，则利用树剖逐渐往上跳
总共要跳$N/pace$次， 一共有$logN$条轻重链， 复杂度为$O(logN+ sqrt{N})$
代码实现比较麻烦， 我常数写的还很差， 水平低啊QAQ
Code
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 #define rd read() 6 #define N 50005 7 #define M 240 8 #define R register 9 using namespace std; 10 11 int n, blo, a[N], b[N], f[N][M], sum[N][M]; 12 int fa[N], dep[N], top[N], sz[N], son[N], id[N], idf[N], cnt; 13 int head[N], tot; 14 15 struct edge { 16 int nxt, to; 17 }e[N << 1]; 18 19 inline char nc(){ 20 static char buf[100000], *p1=buf, *p2=buf; 21 return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++; 22 } 23 inline int read(){ 24 char ch = nc();int sum = 0; 25 while(!(ch >= '0' && ch <= '9')) ch = nc(); 26 while(ch >= '0' && ch <= '9') sum = sum * 10 + ch - 48, ch = nc(); 27 return sum; 28 } 29 30 inline void add(int u, int v) { 31 e[++tot].to = v; 32 e[tot].nxt = head[u]; 33 head[u] = tot; 34 } 35 36 inline void sw(int &A, int &B) { 37 A ^= B; B ^= A; A ^= B; 38 } 39 40 inline void dfs1(R int u) { 41 sz[u] = 1; 42 for (R int i = head[u]; i; i = e[i].nxt) { 43 R int nt = e[i].to; 44 if (nt == fa[u]) continue; 45 f[nt][1] = fa[nt] = u; 46 dep[nt] = dep[u] + 1; 47 dfs1(nt); 48 sz[u] += sz[nt]; 49 if (sz[nt] > sz[son[u]]) 50 son[u] = nt; 51 } 52 } 53 54 inline void dfs2(R int u) { 55 idf[id[u] = ++cnt] = u; 56 if (!son[u]) return; 57 top[son[u]] = top[u]; 58 dfs2(son[u]); 59 for (R int i = head[u]; i; i = e[i].nxt) { 60 R int nt = e[i].to; 61 if (nt == fa[u] || nt == son[u]) continue; 62 top[nt] = nt; 63 dfs2(nt); 64 } 65 } 66 67 inline int LCA(R int x, R int y) { 68 for (;top[x] != top[y]; ) { 69 if (dep[top[x]] < dep[top[y]]) sw(x, y); 70 x = fa[top[x]]; 71 } 72 if (dep[x] < dep[y]) sw(x, y); 73 return y; 74 } 75 76 inline int work1(R int x, R int y, R int pace) { 77 R int lca = LCA(x, y), len = dep[x] + dep[y] - 2 * dep[lca], res = 0; 78 if (len % pace) { 79 res += a[y]; 80 y = f[y][len % pace]; 81 len -= len % pace; 82 } 83 len = dep[x] - dep[lca]; 84 if (len % pace == 0) { 85 res += sum[x][pace] - sum[lca][pace]; 86 res += sum[y][pace] - sum[lca][pace]; 87 res += a[lca]; 88 return res; 89 } 90 R int tmp = f[lca][pace - len % pace]; 91 res += sum[x][pace] - sum[tmp][pace]; 92 if (dep[y] < dep[lca]) return res; 93 len = dep[y] - dep[lca]; 94 tmp = f[lca][pace - len % pace]; 95 res += sum[y][pace] - sum[tmp][pace]; 96 return res; 97 } 98 99 inline int up(R int x, R int d) { 100 R int y = top[x]; 101 for (; x && dep[x] - dep[y] < d;) { 102 d -= dep[x] - dep[y] + 1; 103 x = fa[top[x]]; 104 y = top[x]; 105 } 106 if (!x) return 0; 107 return idf[id[x] - d]; 108 } 109 110 inline int work2(R int x, R int y, R int pace) { 111 R int lca = LCA(x, y), len = dep[x] + dep[y] - 2 * dep[lca], res = 0; 112 if (len % pace) { 113 res += a[y]; 114 y = up(y, len % pace); 115 len -= len % pace; 116 } 117 len = dep[x] - dep[lca]; 118 if (len % pace == 0) { 119 while (x && dep[x] > dep[lca]) 120 res += a[x], x = up(x, pace); 121 while (y && dep[y] > dep[lca]) 122 res += a[y], y = up(y, pace); 123 res += a[lca]; 124 return res; 125 } 126 while (x && dep[x] > dep[lca]) 127 res += a[x], x = up(x, pace); 128 while (y && dep[y] > dep[lca]) 129 res += a[y], y = up(y, pace); 130 return res; 131 } 132 133 int main() 134 { 135 n = rd; blo = sqrt(n); 136 for (R int i = 1; i <= n; ++i) 137 a[i] = rd; 138 for (R int i = 1; i < n; ++i) { 139 int u = rd, v = rd; 140 add(u, v); add(v, u); 141 } 142 dep[1] = 1; dfs1(1); 143 top[1] = 1; dfs2(1); 144 for (R int j = 2; j <= blo; ++j) 145 for (R int i = 1; i <= n; ++i) 146 f[i][j] = f[f[i][j - 1]][1]; 147 for (R int j = 1; j <= blo; ++j) 148 for (R int i = 1; i <= n; ++i) 149 sum[i][j] = a[i]; 150 for (R int j = 1; j <= blo; ++j) 151 for (R int i = 1; i <= n; ++i) { 152 int x = idf[i]; 153 sum[x][j] += sum[f[x][j]][j]; 154 } 155 for (R int i = 1; i <= n; ++i) b[i] = rd; 156 for (R int i = 1; i < n; ++i) { 157 R int x = rd; 158 if (x <= blo) printf("%d ", work1(b[i], b[i + 1], x)); 159 else printf("%d ", work2(b[i], b[i + 1], x)); 160 } 161 }