#include <cstdio> #include <cstdlib> // #include <cassert> #include <vector> static const unsigned int MODULUS = 1000 * 1000 * 1000 + 7; struct node { int parent; int left; int right; unsigned int size; unsigned int left_lin, right_lin; // Linearization cost }; node ltree[500 * 1000 + 1] = { 0 }; node rtree[500 * 1000 + 1] = { 0 }; int stack[500 * 1000 + 1]; int read_tree(node *tree, int n) { int root = -1; for (int i = 1; i <= n; i++) { int parent; scanf("%d", &parent); tree[i].parent = parent; if (parent == -1) { root = i; } if (parent > i) { tree[parent].left = i; } else { tree[parent].right = i; } } return root; } struct visitor_base { bool pre_order(int idx) { return true; } bool in_order(int idx) { return true; } void post_order(int idx) {} }; // TODO: This can be made stackless? template<typename V> void visit_tree(node *tree, int root, V visitor) { int sdepth = 1; stack[0] = root << 2; while (sdepth > 0) { const int curr = stack[sdepth - 1] >> 2; const int progress = stack[sdepth - 1] & 3; if (curr == 0) { sdepth--; continue; } stack[sdepth - 1]++; if (progress == 0) { if (visitor.pre_order(curr)) { stack[sdepth++] = tree[curr].left << 2; } else { sdepth--; } } else if (progress == 1) { if (visitor.in_order(curr)) { stack[sdepth++] = tree[curr].right << 2; } else { sdepth--; } } else { visitor.post_order(curr); sdepth--; } } } void precompute_parameters(node *tree, int root) { struct precomputer : public visitor_base { node *tree; int root; precomputer(node* tree, int root) : visitor_base() , tree(tree) , root(root) {} void post_order(int idx) { tree[idx].size = 1 + tree[tree[idx].left].size + tree[tree[idx].right].size; tree[idx].size %= MODULUS; tree[idx].left_lin = tree[tree[idx].left].left_lin + tree[tree[idx].right].right_lin + tree[tree[idx].right].size; tree[idx].left_lin %= MODULUS; tree[idx].right_lin = tree[tree[idx].right].right_lin + tree[tree[idx].left].left_lin + tree[tree[idx].left].size; tree[idx].right_lin %= MODULUS; } }; visit_tree(tree, root, precomputer{tree, root}); } unsigned int compute_total_cost(int lroot, int rroot) { struct state { int lnode, rnode; int extrude; int dir_from; }; std::vector<state> st; st.push_back(state { lroot, rroot, 0, -1 }); unsigned int ops = 0; while (!st.empty()) { const state s = st.back(); st.pop_back(); int lnode = s.lnode; int rnode = s.rnode; int extrude = s.extrude; const int dir_from = s.dir_from; if (lnode == 0 && rnode == 0) { continue; } int balance, llnode, lrnode, rlnode, rrnode; rlnode = rtree[rnode].left; rrnode = rtree[rnode].right; if (extrude == 0) { llnode = ltree[lnode].left; lrnode = ltree[lnode].right; balance = ltree[llnode].size - rtree[rlnode].size; } else if (dir_from < 0) { llnode = lnode; lrnode = 0; balance = ltree[lnode].size + extrude - 1 - rtree[rlnode].size; } else { llnode = 0; lrnode = lnode; balance = 0 - rtree[rlnode].size; } // printf("At %d %d\n", s.lnode, s.rnode); // printf(" balance: %d\n", balance); // printf(" extrude: %d\n", extrude); // printf(" dir: %d\n", dir_from); // printf(" lsize: %d\n", ltree[lnode].size); // printf(" rsize: %d\n", rtree[rnode].size); // assert(ltree[lnode].size + extrude == rtree[rnode].size); // Now, extrude tells us how many verts we can push to the other side if (extrude > 0) { extrude--; } if (balance > 0 || (balance == 0 && dir_from < 0)) { // lnode // / // / <- extrusion // / // . // / \ // ... ... // Assume that we moved the needed amount of nodes to the right, and compute balance for that // printf(" pushing right %d\n", balance); st.push_back(state { lrnode, rrnode, balance, 1 }); while (extrude < balance) { // Linearize its right subtree, and then move the the left path const int rch = ltree[llnode].right; extrude += 1 + ltree[rch].size; ops = (ops + ltree[rch].right_lin + ltree[rch].size) % MODULUS; // printf(" right-linearizing %d, cost: %d + %d\n", rch, ltree[rch].right_lin, ltree[rch].size); // Advance to the next vertex on the left llnode = ltree[llnode].left; } // Move `extrude` elements from the left path, using rotations, to the right subtree extrude -= balance; ops = (ops + balance) % MODULUS; // Compute left subtree // printf(" pushing left %d\n", extrude); st.push_back(state { llnode, rlnode, extrude, -1 }); } else { // lnode // \ // \ <- extrusion // \ // . // / \ // ... ... balance = -balance; // Assume that we moved the needed amount of nodes to the left, and compute balance for that // printf(" pushing left %d\n", balance); st.push_back(state { llnode, rlnode, balance, -1 }); while (extrude < balance) { // Linearize its left subtree, and then move the the right path const int lch = ltree[lrnode].left; extrude += 1 + ltree[lch].size; ops = (ops + ltree[lch].left_lin + ltree[lch].size) % MODULUS; // printf(" left-linearizing %d, cost: %d + %d\n", lch, ltree[lch].left_lin, ltree[lch].size); // Advance to the next vertex on the right lrnode = ltree[lrnode].right; } // Move `extrude` elements from the right path, using rotations, to the left subtree extrude -= balance; ops = (ops + balance) % MODULUS; // Compute right subtree // printf(" pushing right %d\n", extrude); st.push_back(state { lrnode, rrnode, extrude, 1 }); } // printf(" cost -> %u\n", ops); } return ops; } int main() { int n; scanf("%d", &n); const int lroot = read_tree(ltree, n); const int rroot = read_tree(rtree, n); precompute_parameters(ltree, lroot); precompute_parameters(rtree, rroot); unsigned int result = compute_total_cost(lroot, rroot); printf("%u\n", result); return 0; }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 | #include <cstdio> #include <cstdlib> // #include <cassert> #include <vector> static const unsigned int MODULUS = 1000 * 1000 * 1000 + 7; struct node { int parent; int left; int right; unsigned int size; unsigned int left_lin, right_lin; // Linearization cost }; node ltree[500 * 1000 + 1] = { 0 }; node rtree[500 * 1000 + 1] = { 0 }; int stack[500 * 1000 + 1]; int read_tree(node *tree, int n) { int root = -1; for (int i = 1; i <= n; i++) { int parent; scanf("%d", &parent); tree[i].parent = parent; if (parent == -1) { root = i; } if (parent > i) { tree[parent].left = i; } else { tree[parent].right = i; } } return root; } struct visitor_base { bool pre_order(int idx) { return true; } bool in_order(int idx) { return true; } void post_order(int idx) {} }; // TODO: This can be made stackless? template<typename V> void visit_tree(node *tree, int root, V visitor) { int sdepth = 1; stack[0] = root << 2; while (sdepth > 0) { const int curr = stack[sdepth - 1] >> 2; const int progress = stack[sdepth - 1] & 3; if (curr == 0) { sdepth--; continue; } stack[sdepth - 1]++; if (progress == 0) { if (visitor.pre_order(curr)) { stack[sdepth++] = tree[curr].left << 2; } else { sdepth--; } } else if (progress == 1) { if (visitor.in_order(curr)) { stack[sdepth++] = tree[curr].right << 2; } else { sdepth--; } } else { visitor.post_order(curr); sdepth--; } } } void precompute_parameters(node *tree, int root) { struct precomputer : public visitor_base { node *tree; int root; precomputer(node* tree, int root) : visitor_base() , tree(tree) , root(root) {} void post_order(int idx) { tree[idx].size = 1 + tree[tree[idx].left].size + tree[tree[idx].right].size; tree[idx].size %= MODULUS; tree[idx].left_lin = tree[tree[idx].left].left_lin + tree[tree[idx].right].right_lin + tree[tree[idx].right].size; tree[idx].left_lin %= MODULUS; tree[idx].right_lin = tree[tree[idx].right].right_lin + tree[tree[idx].left].left_lin + tree[tree[idx].left].size; tree[idx].right_lin %= MODULUS; } }; visit_tree(tree, root, precomputer{tree, root}); } unsigned int compute_total_cost(int lroot, int rroot) { struct state { int lnode, rnode; int extrude; int dir_from; }; std::vector<state> st; st.push_back(state { lroot, rroot, 0, -1 }); unsigned int ops = 0; while (!st.empty()) { const state s = st.back(); st.pop_back(); int lnode = s.lnode; int rnode = s.rnode; int extrude = s.extrude; const int dir_from = s.dir_from; if (lnode == 0 && rnode == 0) { continue; } int balance, llnode, lrnode, rlnode, rrnode; rlnode = rtree[rnode].left; rrnode = rtree[rnode].right; if (extrude == 0) { llnode = ltree[lnode].left; lrnode = ltree[lnode].right; balance = ltree[llnode].size - rtree[rlnode].size; } else if (dir_from < 0) { llnode = lnode; lrnode = 0; balance = ltree[lnode].size + extrude - 1 - rtree[rlnode].size; } else { llnode = 0; lrnode = lnode; balance = 0 - rtree[rlnode].size; } // printf("At %d %d\n", s.lnode, s.rnode); // printf(" balance: %d\n", balance); // printf(" extrude: %d\n", extrude); // printf(" dir: %d\n", dir_from); // printf(" lsize: %d\n", ltree[lnode].size); // printf(" rsize: %d\n", rtree[rnode].size); // assert(ltree[lnode].size + extrude == rtree[rnode].size); // Now, extrude tells us how many verts we can push to the other side if (extrude > 0) { extrude--; } if (balance > 0 || (balance == 0 && dir_from < 0)) { // lnode // / // / <- extrusion // / // . // / \ // ... ... // Assume that we moved the needed amount of nodes to the right, and compute balance for that // printf(" pushing right %d\n", balance); st.push_back(state { lrnode, rrnode, balance, 1 }); while (extrude < balance) { // Linearize its right subtree, and then move the the left path const int rch = ltree[llnode].right; extrude += 1 + ltree[rch].size; ops = (ops + ltree[rch].right_lin + ltree[rch].size) % MODULUS; // printf(" right-linearizing %d, cost: %d + %d\n", rch, ltree[rch].right_lin, ltree[rch].size); // Advance to the next vertex on the left llnode = ltree[llnode].left; } // Move `extrude` elements from the left path, using rotations, to the right subtree extrude -= balance; ops = (ops + balance) % MODULUS; // Compute left subtree // printf(" pushing left %d\n", extrude); st.push_back(state { llnode, rlnode, extrude, -1 }); } else { // lnode // \ // \ <- extrusion // \ // . // / \ // ... ... balance = -balance; // Assume that we moved the needed amount of nodes to the left, and compute balance for that // printf(" pushing left %d\n", balance); st.push_back(state { llnode, rlnode, balance, -1 }); while (extrude < balance) { // Linearize its left subtree, and then move the the right path const int lch = ltree[lrnode].left; extrude += 1 + ltree[lch].size; ops = (ops + ltree[lch].left_lin + ltree[lch].size) % MODULUS; // printf(" left-linearizing %d, cost: %d + %d\n", lch, ltree[lch].left_lin, ltree[lch].size); // Advance to the next vertex on the right lrnode = ltree[lrnode].right; } // Move `extrude` elements from the right path, using rotations, to the left subtree extrude -= balance; ops = (ops + balance) % MODULUS; // Compute right subtree // printf(" pushing right %d\n", extrude); st.push_back(state { lrnode, rrnode, extrude, 1 }); } // printf(" cost -> %u\n", ops); } return ops; } int main() { int n; scanf("%d", &n); const int lroot = read_tree(ltree, n); const int rroot = read_tree(rtree, n); precompute_parameters(ltree, lroot); precompute_parameters(rtree, rroot); unsigned int result = compute_total_cost(lroot, rroot); printf("%u\n", result); return 0; } |