这两天我都在干嘛= =。。。浪死了
啊啊啊终于调出来了这道2b题。。。
莫队~莫队~但是注意要直接树分块！
按L排序，分块R和Change即可
具体方法还有复杂度什么的详见vfk的blog好了
1 /************************************************************** 2 Problem: 3052 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:69886 ms 7 Memory:29368 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 14 using namespace std; 15 typedef long long ll; 16 typedef double lf; 17 const int N = 100005; 18 const int Maxlen = N * 60; 19 20 ll now, ans[N]; 21 int n, m, Qu; 22 int cnt_block, sz_block; 23 ll v[N], w[N], pre[N]; 24 int cnt_c[N]; 25 int q[N], top_q; 26 27 char buf[Maxlen], *c = buf; 28 int Len; 29 30 struct tree_node { 31 int sz, dep, dfn, w; 32 int fa[17]; 33 ll c; 34 bool vis; 35 } tr[N]; 36 37 int cnt_tree; 38 39 struct edge { 40 int next, to; 41 edge() {} 42 edge(int _n, int _t) : next(_n), to(_t) {} 43 } e[N << 1]; 44 45 int first[N], cnt_edge; 46 47 struct oper_query { 48 int x, y, id, t; 49 oper_query() {} 50 oper_query(int _x, int _y, int _i, int _t) : x(_x), y(_y), id(_i), t(_t) {} 51 52 inline bool operator < (const oper_query &X) const { 53 return tr[x].w != tr[X.x].w ? tr[x].w < tr[X.x].w : 54 tr[y].w != tr[X.y].w ? tr[y].w < tr[X.y].w : t < X.t; 55 } 56 } oq[N]; 57 58 int cnt_query; 59 60 struct oper_change { 61 int x, y, pre; 62 oper_change() {} 63 oper_change(int _x, int _y, int _p) : x(_x), y(_y), pre(_p) {} 64 } oc[N]; 65 66 int cnt_change; 67 68 inline int read() { 69 int x = 0; 70 while (*c < '0' || '9' < *c) ++c; 71 while ('0' <= *c && *c <= '9') 72 x = x * 10 + *c - '0', ++c; 73 return x; 74 } 75 76 inline void Add_Edges(int x, int y) { 77 e[++cnt_edge] = edge(first[x], y), first[x] = cnt_edge; 78 e[++cnt_edge] = edge(first[y], x), first[y] = cnt_edge; 79 } 80 81 void dfs(int p) { 82 int i, x, y; 83 tr[p].dfn = ++cnt_tree; 84 for (i = 1; i <= 16; ++i) 85 tr[p].fa[i] = tr[tr[p].fa[i - 1]].fa[i - 1]; 86 for (x = first[p]; x; x = e[x].next) 87 if ((y = e[x].to) != tr[p].fa[0]) { 88 tr[y].dep = tr[p].dep + 1, tr[y].fa[0] = p; 89 dfs(y); 90 tr[p].sz += tr[y].sz; 91 if (tr[p].sz >= sz_block) { 92 for (i = 1, ++cnt_block; i <= tr[p].sz; ++i) 93 tr[q[top_q--]].w = cnt_block; 94 tr[p].sz = 0; 95 } 96 } 97 q[++top_q] = p; 98 ++tr[p].sz; 99 } 100 101 inline int lca(int x, int y) { 102 int i; 103 if (tr[x].dep < tr[y].dep) swap(x, y); 104 for (i = 16; ~i; --i) 105 if (tr[tr[x].fa[i]].dep >= tr[y].dep) 106 x = tr[x].fa[i]; 107 if (x == y) return x; 108 for (i = 16; ~i; --i) 109 if (tr[x].fa[i] != tr[y].fa[i]) 110 x = tr[x].fa[i], y = tr[y].fa[i]; 111 return tr[x].fa[0]; 112 } 113 114 115 inline void reverse(int p) { 116 if (tr[p].vis) 117 now -= w[cnt_c[tr[p].c]--] * v[tr[p].c]; 118 else 119 now += w[++cnt_c[tr[p].c]] * v[tr[p].c]; 120 tr[p].vis ^= 1; 121 } 122 123 inline void update(int p, int C) { 124 if (tr[p].vis) { 125 reverse(p); 126 tr[p].c = C; 127 reverse(p); 128 } else tr[p].c = C; 129 } 130 131 inline void solve(int x, int y) { 132 while (x != y) { 133 if (tr[x].dep > tr[y].dep) 134 reverse(x), x = tr[x].fa[0]; 135 else reverse(y), y = tr[y].fa[0]; 136 } 137 } 138 139 int main() { 140 int i, j, LCA, oper, x, y, tot_Q; 141 Len = fread(c, 1, Maxlen, stdin); 142 buf[Len] = ' '; 143 n = read(), m = read(), tot_Q = read(); 144 sz_block = (int) pow(n, (lf) 2.0 / 3) / 2; 145 for (i = 1; i <= m; ++i) 146 v[i] = read(); 147 for (i = 1; i <= n; ++i) 148 w[i] = read(); 149 for (i = 1; i < n; ++i) 150 Add_Edges(read(), read()); 151 for (i = 1; i <= n; ++i) 152 pre[i] = tr[i].c = read(); 153 154 tr[1].dep = 1; 155 dfs(1); 156 while (top_q) 157 tr[q[top_q--]].w = cnt_block; 158 159 for (i = 1; i <= tot_Q; ++i) { 160 oper = read(), x = read(), y = read(); 161 if (!oper) 162 oc[++cnt_change] = oper_change(x, y, pre[x]), pre[x] = y; 163 else { 164 if (tr[x].dfn > tr[y].dfn) swap(x, y); 165 oq[++cnt_query] = oper_query(x, y, cnt_query, cnt_change); 166 } 167 } 168 169 sort(oq + 1, oq + cnt_query + 1); 170 for (i = 1, oq[0].t = 0; i <= cnt_query; ++i) { 171 for (j = oq[i - 1].t + 1; j <= oq[i].t; ++j) 172 update(oc[j].x, oc[j].y); 173 for (j = oq[i - 1].t; j > oq[i].t; --j) 174 update(oc[j].x, oc[j].pre); 175 if (i == 1) solve(oq[i].x, oq[i].y); 176 else solve(oq[i - 1].x, oq[i].x), solve(oq[i - 1].y, oq[i].y); 177 LCA = lca(oq[i].x, oq[i].y); 178 reverse(LCA), ans[oq[i].id] = now, reverse(LCA); 179 } 180 for (i = 1; i <= cnt_query; ++i) 181 printf("%lld ", ans[i]); 182 return 0; 183 }