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| 1 | +package weekly; |
| 2 | + |
| 3 | +import java.util.ArrayList; |
| 4 | +import java.util.Arrays; |
| 5 | +import java.util.HashMap; |
| 6 | +import java.util.List; |
| 7 | +import java.util.Map; |
| 8 | + |
| 9 | +public class wk361 { |
| 10 | + |
| 11 | + //枚举 |
| 12 | + public int countSymmetricIntegers(int low, int high) { |
| 13 | + |
| 14 | + int ans = 0; |
| 15 | + for (int i = low; i <= high; i++) { |
| 16 | + int num = i; |
| 17 | + List<Integer> list = new ArrayList<>(); |
| 18 | + while (num > 0) { |
| 19 | + list.add(num % 10); |
| 20 | + num /= 10; |
| 21 | + } |
| 22 | + if (list.size() % 2 == 1) continue; |
| 23 | + int sum = 0; |
| 24 | + for (int j = 0; j < list.size() / 2; j++) { |
| 25 | + sum += list.get(j); |
| 26 | + } |
| 27 | + for (int j = list.size() / 2; j < list.size(); j++) { |
| 28 | + sum -= list.get(j); |
| 29 | + } |
| 30 | + if (sum == 0) ans++; |
| 31 | + } |
| 32 | + return ans; |
| 33 | + } |
| 34 | + |
| 35 | + |
| 36 | + //枚举后缀 |
| 37 | + public int minimumOperations(String num) { |
| 38 | + int ans = num.length(); |
| 39 | + //变成0 |
| 40 | + if (num.contains("0")) { |
| 41 | + ans = num.length() - 1; |
| 42 | + } |
| 43 | + |
| 44 | + ans = Math.min(ans, help(num, new char[]{'5', '2'})); |
| 45 | + ans = Math.min(ans, help(num, new char[]{'0', '5'})); |
| 46 | + ans = Math.min(ans, help(num, new char[]{'5', '7'})); |
| 47 | + ans = Math.min(ans, help(num, new char[]{'0', '0'})); |
| 48 | + return ans; |
| 49 | + } |
| 50 | + |
| 51 | + int help(String num, char[] arr) { |
| 52 | + int index = 0; |
| 53 | + int ans = 0; |
| 54 | + for (int i = num.length() - 1; i >= 0; i--) { |
| 55 | + if (num.charAt(i) == arr[index]) { |
| 56 | + index++; |
| 57 | + if (index >= arr.length) break; |
| 58 | + } else { |
| 59 | + ans++; |
| 60 | + } |
| 61 | + } |
| 62 | + if (index < arr.length) return Integer.MAX_VALUE; |
| 63 | + return ans; |
| 64 | + } |
| 65 | + |
| 66 | + |
| 67 | + //前缀和 |
| 68 | + static public long countInterestingSubarrays(List<Integer> nums, int modulo, int k) { |
| 69 | + int[] pre = new int[nums.size() + 1]; |
| 70 | + for (int i = 0; i < nums.size(); i++) { |
| 71 | + pre[i + 1] = pre[i] + (nums.get(i) % modulo == k ? 1 : 0); |
| 72 | + } |
| 73 | + |
| 74 | + Map<Integer, Integer> map = new HashMap<>(); |
| 75 | + |
| 76 | + long ans = 0; |
| 77 | + for (int i = 0; i < pre.length; i++) { |
| 78 | + int num = pre[i]; |
| 79 | + ans += map.getOrDefault((num %modulo - k + modulo) % modulo,0); |
| 80 | + map.put(num % modulo, map.getOrDefault(num % modulo, 0) + 1); |
| 81 | + } |
| 82 | + return ans; |
| 83 | + } |
| 84 | + |
| 85 | + |
| 86 | + /* static public long countInterestingSubarrays(List<Integer> nums, int modulo, int k) { |
| 87 | + int[][] check = new int[nums.size()][2]; |
| 88 | + int pre = 0; |
| 89 | + for (int i = 0; i < nums.size(); i++) { |
| 90 | + int num = nums.get(i); |
| 91 | + if (num % modulo == k) { |
| 92 | + check[i][0] = 1; |
| 93 | + check[i][1] = pre; |
| 94 | + pre = 0; |
| 95 | + } else { |
| 96 | + pre++; |
| 97 | + } |
| 98 | + } |
| 99 | +
|
| 100 | + List<int[]> list = new ArrayList<>(); |
| 101 | + long ans = 0; |
| 102 | + pre = 0; |
| 103 | + for (int i = 0; i < nums.size(); i++) { |
| 104 | + int sum = 0; |
| 105 | + if (check[i][0] > 0) { |
| 106 | + int left = list.size() - modulo; |
| 107 | + list.add(new int[]{i, (left >= 0 && left < list.size() ? list.get(left)[1] : 0) + pre + 1}); |
| 108 | + pre = 0; |
| 109 | + } else { |
| 110 | + pre++; |
| 111 | + if (k == 0) sum = pre; |
| 112 | + } |
| 113 | + int j = k == 0 ? list.size() - modulo : list.size() - k; |
| 114 | + if (j >= 0) sum += list.get(j)[1]; |
| 115 | + ans += sum; |
| 116 | + } |
| 117 | + return ans; |
| 118 | + }*/ |
| 119 | + |
| 120 | + |
| 121 | + |
| 122 | + //双上倍增+lca |
| 123 | + public int[] minOperationsQueries(int n, int[][] edges, int[][] queries) { |
| 124 | + List<int[]>[] g = new ArrayList[n]; |
| 125 | + Arrays.setAll(g, e -> new ArrayList<>()); |
| 126 | + for (int [] e : edges) { |
| 127 | + int x = e[0], y = e[1], w = e[2] - 1; |
| 128 | + g[x].add(new int[]{y, w}); |
| 129 | + g[y].add(new int[]{x, w}); |
| 130 | + } |
| 131 | + |
| 132 | + int m = 32 - Integer.numberOfLeadingZeros(n); // n 的二进制长度 |
| 133 | + int [][] pa = new int[n][m]; |
| 134 | + for (int i = 0; i < n; i++) { |
| 135 | + Arrays.fill(pa[i], -1); |
| 136 | + } |
| 137 | + int [][][] cnt = new int[n][m][26]; |
| 138 | + int [] depth = new int[n]; |
| 139 | + //求出 父节点、cnx[x][0]，深度数组 |
| 140 | + dfs(0, -1, g, pa, cnt, depth); |
| 141 | + |
| 142 | + //倍增求每个点的2的i次方的次数以及位置 |
| 143 | + for (int i = 0; i < m - 1; i++) { |
| 144 | + for (int x = 0; x < n; x++) { |
| 145 | + int p = pa[x][i]; |
| 146 | + if (p != -1) { |
| 147 | + int pp = pa[p][i]; |
| 148 | + pa[x][i + 1] = pp; |
| 149 | + for (int j = 0; j < 26; j++) { |
| 150 | + cnt[x][i + 1][j] = cnt[x][i][j] + cnt[p][i][j]; |
| 151 | + } |
| 152 | + } |
| 153 | + } |
| 154 | + } |
| 155 | + |
| 156 | + //计算结果 |
| 157 | + int [] ans = new int[queries.length]; |
| 158 | + for (int qi = 0; qi < queries.length; qi++) { |
| 159 | + int x = queries[qi][0], y = queries[qi][1]; |
| 160 | + int pathLen = depth[x] + depth[y]; |
| 161 | + //倍增求lca，顺便计算次数 |
| 162 | + int [] cw = new int[26]; |
| 163 | + //置换深度大小，让y始终是深的 |
| 164 | + if (depth[x] > depth[y]) { |
| 165 | + int temp = x; |
| 166 | + x = y; |
| 167 | + y = temp; |
| 168 | + } |
| 169 | + |
| 170 | + // 让 y 和 x 在同一深度，避免在同一个路径上 |
| 171 | + for (int k = depth[y] - depth[x]; k > 0; k &= k - 1) { |
| 172 | + int i = Integer.numberOfTrailingZeros(k); |
| 173 | + int p = pa[y][i]; |
| 174 | + for (int j = 0; j < 26; ++j) { |
| 175 | + cw[j] += cnt[y][i][j]; |
| 176 | + } |
| 177 | + y = p; |
| 178 | + } |
| 179 | + if (y != x) { |
| 180 | + for (int i = m - 1; i >= 0; i--) { |
| 181 | + int px = pa[x][i]; |
| 182 | + int py = pa[y][i]; |
| 183 | + //相等表示跳过了 |
| 184 | + if (px != py) { |
| 185 | + for (int j = 0; j < 26; j++) { |
| 186 | + cw[j] += cnt[x][i][j] + cnt[y][i][j]; |
| 187 | + } |
| 188 | + x = px; |
| 189 | + y = py; // x 和 y 同时上跳 2^i 步 |
| 190 | + } |
| 191 | + } |
| 192 | + for (int j = 0; j < 26; j++) { |
| 193 | + cw[j] += cnt[x][0][j] + cnt[y][0][j]; |
| 194 | + } |
| 195 | + x = pa[x][0]; |
| 196 | + } |
| 197 | + |
| 198 | + int lca = x; |
| 199 | + pathLen -= depth[lca] * 2; |
| 200 | + int maxCw = 0; |
| 201 | + for (int i = 0; i < 26; i++) { |
| 202 | + maxCw = Math.max(maxCw, cw[i]); |
| 203 | + } |
| 204 | + ans[qi] = pathLen - maxCw; |
| 205 | + } |
| 206 | + return ans; |
| 207 | + } |
| 208 | + |
| 209 | + private void dfs(int x, int fa, List<int[]>[] g, int[][] pa, int[][][] cnt, int[] depth) { |
| 210 | + pa[x][0] = fa; |
| 211 | + for (int [] e : g[x]) { |
| 212 | + int y = e[0], w = e[1]; |
| 213 | + if (y != fa) { |
| 214 | + cnt[y][0][w] = 1; |
| 215 | + depth[y] = depth[x] + 1; |
| 216 | + dfs(y, x, g, pa, cnt, depth); |
| 217 | + } |
| 218 | + } |
| 219 | + } |
| 220 | + |
| 221 | + |
| 222 | +} |
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