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| 1 | +package weekly; |
| 2 | + |
| 3 | +import java.util.ArrayDeque; |
| 4 | +import java.util.ArrayList; |
| 5 | +import java.util.Arrays; |
| 6 | +import java.util.Comparator; |
| 7 | +import java.util.HashSet; |
| 8 | +import java.util.LinkedList; |
| 9 | +import java.util.List; |
| 10 | +import java.util.PriorityQueue; |
| 11 | +import java.util.Queue; |
| 12 | +import java.util.Set; |
| 13 | +import java.util.TreeSet; |
| 14 | + |
| 15 | +public class wk339 { |
| 16 | + |
| 17 | + //模拟 |
| 18 | + static public int findTheLongestBalancedSubstring(String s) { |
| 19 | + int i = 0; |
| 20 | + int ans = 0; |
| 21 | + while (i < s.length()) { |
| 22 | + int zero = 0; |
| 23 | + while (i < s.length() && s.charAt(i) == '0') { |
| 24 | + zero++; |
| 25 | + i++; |
| 26 | + } |
| 27 | + int one = 0; |
| 28 | + while (i < s.length() && s.charAt(i) == '1') { |
| 29 | + one++; |
| 30 | + i++; |
| 31 | + } |
| 32 | + ans = Math.max(Math.min(zero, one) * 2, ans); |
| 33 | + } |
| 34 | + return ans; |
| 35 | + } |
| 36 | + |
| 37 | + |
| 38 | + //hash |
| 39 | + public List<List<Integer>> findMatrix(int[] nums) { |
| 40 | + Arrays.sort(nums); |
| 41 | + |
| 42 | + int pre = nums[0]; |
| 43 | + int count = 1; |
| 44 | + List<List<Integer>> res = new ArrayList<>(); |
| 45 | + for (int i = 1; i < nums.length; i++) { |
| 46 | + if (nums[i] == pre) { |
| 47 | + count++; |
| 48 | + } else { |
| 49 | + while (res.size() < count) { |
| 50 | + res.add(new ArrayList<>()); |
| 51 | + } |
| 52 | + for (int j = 0; j < count; j++) { |
| 53 | + res.get(j).add(pre); |
| 54 | + } |
| 55 | + pre = nums[i]; |
| 56 | + count = 1; |
| 57 | + } |
| 58 | + } |
| 59 | + while (res.size() < count) { |
| 60 | + res.add(new ArrayList<>()); |
| 61 | + } |
| 62 | + for (int j = 0; j < count; j++) { |
| 63 | + res.get(j).add(pre); |
| 64 | + } |
| 65 | + |
| 66 | + return res; |
| 67 | + } |
| 68 | + |
| 69 | + |
| 70 | + //贪心 |
| 71 | + static public int miceAndCheese(int[] reward1, int[] reward2, int k) { |
| 72 | + |
| 73 | + PriorityQueue<Integer> priorityQueue = new PriorityQueue<>(Comparator.reverseOrder()); |
| 74 | + |
| 75 | + int sum = 0; |
| 76 | + for (int i = 0; i < reward1.length; i++) { |
| 77 | + int dis = reward1[i] - reward2[i]; |
| 78 | + priorityQueue.add(dis); |
| 79 | + sum += reward2[i]; |
| 80 | + } |
| 81 | + |
| 82 | + while (k > 0) { |
| 83 | + sum += priorityQueue.poll(); |
| 84 | + k--; |
| 85 | + } |
| 86 | + return sum; |
| 87 | + |
| 88 | + } |
| 89 | + |
| 90 | + |
| 91 | + //直接遍历超时 |
| 92 | + /*static public int[] minReverseOperations(int n, int p, int[] banned, int k) { |
| 93 | + int[] res = new int[n]; |
| 94 | + Set<Integer> banSet = new HashSet<>(); |
| 95 | + for (int i : banned) { |
| 96 | + banSet.add(i); |
| 97 | + } |
| 98 | + //存储未查询的点 |
| 99 | + Set<Integer> treeSet=new TreeSet<>(); |
| 100 | + for (int i = 0; i < n; i++) { |
| 101 | + treeSet.add(i); |
| 102 | + } |
| 103 | + Queue<Integer> queue = new LinkedList<>(); |
| 104 | + queue.add(p); |
| 105 | + treeSet.remove(p); |
| 106 | + for (int i = 0; i < res.length; i++) { |
| 107 | + res[i] = -1; |
| 108 | + } |
| 109 | + res[p] = 0; |
| 110 | + boolean[] visited = new boolean[n]; |
| 111 | + int step = 0; |
| 112 | + while (!queue.isEmpty()) { |
| 113 | + step++; |
| 114 | + int size = queue.size(); |
| 115 | + while (size-- > 0) { |
| 116 | + int cur = queue.poll(); |
| 117 | + visited[cur] = true; |
| 118 | +
|
| 119 | + int r=cur+k-1; |
| 120 | + int l=cur-k+1; |
| 121 | + for (int i = k; i > 1; i -= 2) { |
| 122 | + //左移 |
| 123 | + int left = cur - i + 1; |
| 124 | + int ll = (l++); |
| 125 | + int lr = ll+k-1; |
| 126 | + //右移动 |
| 127 | + int right = cur + i - 1; |
| 128 | + int rr=(r--); |
| 129 | + int rl=rr-k+1; |
| 130 | +
|
| 131 | + if (left >= 0 && lr<n&&ll>=0 && !visited[left] && !banSet.contains(left)) { |
| 132 | + visited[left] = true; |
| 133 | + queue.add(left); |
| 134 | + res[left] = step; |
| 135 | + } |
| 136 | +
|
| 137 | + if (right < n &&rl>=0&&rr<n && !visited[right] && !banSet.contains(right)) { |
| 138 | + visited[right] = true; |
| 139 | + queue.add(right); |
| 140 | + res[right] = step; |
| 141 | + } |
| 142 | + } |
| 143 | + } |
| 144 | + } |
| 145 | + return res; |
| 146 | + }*/ |
| 147 | + |
| 148 | + |
| 149 | + //广度优先+有序集合 |
| 150 | + static public int[] minReverseOperations(int n, int p, int[] banned, int k) { |
| 151 | + int[] res = new int[n]; |
| 152 | + Set<Integer> banSet = new HashSet<>(); |
| 153 | + for (int i : banned) { |
| 154 | + banSet.add(i); |
| 155 | + } |
| 156 | + banSet.add(p); |
| 157 | + //存储未查询的点 |
| 158 | + TreeSet<Integer> treeEven = new TreeSet<>(); |
| 159 | + TreeSet<Integer> treeOdd = new TreeSet<>(); |
| 160 | + |
| 161 | + TreeSet<Integer> sets[] = new TreeSet[2]; |
| 162 | + sets[0] = treeEven; |
| 163 | + sets[1] = treeOdd; |
| 164 | + treeEven.add(n); |
| 165 | + treeOdd.add(n); |
| 166 | + for (int i = 0; i < n; i++) { |
| 167 | + if (banSet.contains(i)) continue; |
| 168 | + if (i % 2 == 0) { |
| 169 | + treeEven.add(i); |
| 170 | + } else { |
| 171 | + treeOdd.add(i); |
| 172 | + } |
| 173 | + } |
| 174 | + Queue<Integer> queue = new LinkedList<>(); |
| 175 | + queue.add(p); |
| 176 | + for (int i = 0; i < res.length; i++) { |
| 177 | + res[i] = -1; |
| 178 | + } |
| 179 | + int step = 0; |
| 180 | + while (!queue.isEmpty()) { |
| 181 | + int size = queue.size(); |
| 182 | + while (size-- > 0) { |
| 183 | + Integer i = queue.poll(); |
| 184 | + res[i] = step; |
| 185 | + // 从 mn 到 mx 的所有位置都可以翻转到 |
| 186 | + int mn = Math.max(i - k + 1, k - i - 1); |
| 187 | + int mx = Math.min(i + k - 1, n * 2 - k - i - 1); |
| 188 | + TreeSet<Integer> s = sets[mn % 2]; |
| 189 | + for (int j = s.ceiling(mn); j <= mx; j = s.ceiling(mn)) { |
| 190 | + queue.add(j); |
| 191 | + s.remove(j); |
| 192 | + } |
| 193 | + } |
| 194 | + step++; |
| 195 | + } |
| 196 | + return res; |
| 197 | + } |
| 198 | + |
| 199 | + public static void main(String[] args) { |
| 200 | + minReverseOperations(4, 0, new int[]{1,2}, 4); |
| 201 | + } |
| 202 | +} |
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