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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.HashSet; |
| 7 | +import java.util.List; |
| 8 | +import java.util.Map; |
| 9 | +import java.util.Set; |
| 10 | +import java.util.TreeSet; |
| 11 | + |
| 12 | +public class wk320 { |
| 13 | + |
| 14 | + //ranking: 71 / 5678 |
| 15 | + |
| 16 | + // a*b*c |
| 17 | + public int unequalTriplets(int[] nums) { |
| 18 | + Arrays.sort(nums); |
| 19 | + int start = 0; |
| 20 | + int ans = 0; |
| 21 | + for (int i = 0; i < nums.length - 1; i++) { |
| 22 | + if (nums[i] != nums[i + 1]) { |
| 23 | + ans += start * (i - start + 1) * (nums.length - i - 1); |
| 24 | + start = i + 1; |
| 25 | + } |
| 26 | + } |
| 27 | + return ans; |
| 28 | + } |
| 29 | + |
| 30 | + /* public int unequalTriplets(int[] nums) { |
| 31 | + int res = 0; |
| 32 | + for (int i = 0; i < nums.length; i++) { |
| 33 | + for (int j = i + 1; j < nums.length; j++) { |
| 34 | + for (int k = j + 1; k < nums.length; k++) { |
| 35 | + if (nums[i] != nums[j] && nums[j] != nums[k] && nums[i] != nums[k]) { |
| 36 | + res++; |
| 37 | + } |
| 38 | + } |
| 39 | + } |
| 40 | + } |
| 41 | + return res; |
| 42 | + }*/ |
| 43 | + |
| 44 | + |
| 45 | + // 用treemap做快一点 |
| 46 | + public class TreeNode { |
| 47 | + int val; |
| 48 | + TreeNode left; |
| 49 | + TreeNode right; |
| 50 | + |
| 51 | + TreeNode() { |
| 52 | + } |
| 53 | + |
| 54 | + TreeNode(int val) { |
| 55 | + this.val = val; |
| 56 | + } |
| 57 | + |
| 58 | + TreeNode(int val, TreeNode left, TreeNode right) { |
| 59 | + this.val = val; |
| 60 | + this.left = left; |
| 61 | + this.right = right; |
| 62 | + } |
| 63 | + } |
| 64 | + |
| 65 | + TreeSet<Integer> set = new TreeSet<>(); |
| 66 | + |
| 67 | + public List<List<Integer>> closestNodes(TreeNode root, List<Integer> queries) { |
| 68 | + dfs(root); |
| 69 | + List<List<Integer>> res = new ArrayList<>(); |
| 70 | + for (int i = 0; i < queries.size(); i++) { |
| 71 | + List<Integer> l = new ArrayList<>(); |
| 72 | + Integer floor = set.floor(queries.get(i)); |
| 73 | + if (floor == null) { |
| 74 | + l.add(-1); |
| 75 | + } else { |
| 76 | + l.add(floor); |
| 77 | + } |
| 78 | + Integer ceiling = set.ceiling(queries.get(i)); |
| 79 | + if (ceiling == null) { |
| 80 | + l.add(-1); |
| 81 | + } else { |
| 82 | + l.add(ceiling); |
| 83 | + } |
| 84 | + res.add(l); |
| 85 | + } |
| 86 | + return res; |
| 87 | + } |
| 88 | + |
| 89 | + void dfs(TreeNode root) { |
| 90 | + if (root == null) { |
| 91 | + return; |
| 92 | + } |
| 93 | + set.add(root.val); |
| 94 | + dfs(root.left); |
| 95 | + dfs(root.right); |
| 96 | + } |
| 97 | + |
| 98 | + //贪心 每次把所有乘客都放到离0点更近的节点,一起运送 |
| 99 | + public long minimumFuelCost(int[][] roads, int seats) { |
| 100 | + Map<Integer, List<Integer>> map = new HashMap<>(); |
| 101 | + for (int[] road : roads) { |
| 102 | + if (!map.containsKey(road[0])) map.put(road[0], new ArrayList<>()); |
| 103 | + if (!map.containsKey(road[1])) map.put(road[1], new ArrayList<>()); |
| 104 | + map.get(road[0]).add(road[1]); |
| 105 | + map.get(road[1]).add(road[0]); |
| 106 | + } |
| 107 | + dfs(0, -1, seats, map); |
| 108 | + return cost; |
| 109 | + } |
| 110 | + |
| 111 | + long cost = 0; |
| 112 | + |
| 113 | + int dfs(int cur, int pre, int seats, Map<Integer, List<Integer>> map) { |
| 114 | + |
| 115 | + int count = 0; |
| 116 | + for (Integer next : map.getOrDefault(cur, new ArrayList<>())) { |
| 117 | + if (next == pre) continue; |
| 118 | + int c = dfs(next, cur, seats, map); |
| 119 | + cost += c / seats + (c % seats == 0 ? 0 : 1); |
| 120 | + count += c; |
| 121 | + } |
| 122 | + return count + 1; |
| 123 | + } |
| 124 | + |
| 125 | + |
| 126 | + //dp[i][k] 记录截止到i,能分成k段的方法次数 |
| 127 | + /* static public int beautifulPartitions(String s, int K, int minLength) { |
| 128 | + Set<Character> set = new HashSet<>(Arrays.asList('2', '3', '5', '7')); |
| 129 | + long[][] dp = new long[s.length() + 1][K + 1]; |
| 130 | + dp[0][0] = 1; |
| 131 | + int mod = (int) 1e9 + 7; |
| 132 | + //截止到i |
| 133 | + for (int i = 0; i < s.length(); i++) { |
| 134 | + //从j开始 |
| 135 | + for (int j = i - minLength + 1; j >= 0; j--) { |
| 136 | + //首尾判断 |
| 137 | + if (set.contains(s.charAt(j)) && !set.contains(s.charAt(i))) { |
| 138 | + for (int k = 1; k <= K; k++) { |
| 139 | + dp[i + 1][k] += dp[j][k - 1]; |
| 140 | + dp[i + 1][k] %= mod; |
| 141 | + } |
| 142 | + } |
| 143 | + } |
| 144 | + } |
| 145 | + return (int) dp[s.length()][K]; |
| 146 | + }*/ |
| 147 | + |
| 148 | + //优化成二次循环 |
| 149 | + //dp[k][i] 记录截止到i,能分成k段的方法次数 |
| 150 | + public int beautifulPartitions(String s, int K, int minLength) { |
| 151 | + long[][] dp = new long[s.length() + 1][K + 1]; |
| 152 | + dp[0][0] = 1; |
| 153 | + int mod = (int) 1e9 + 7; |
| 154 | + if (K * minLength > s.length() || !isPrime(s.charAt(0)) || isPrime(s.charAt(s.length() - 1))) // 剪枝 |
| 155 | + return 0; |
| 156 | + //分成k段 |
| 157 | + for (int i = 1; i <= K; i++) { |
| 158 | + long sum = 0; |
| 159 | + //从j开始 |
| 160 | + for (int j = i * minLength; j + (K - i) * minLength <= s.length(); j++) { |
| 161 | + //求前缀和 |
| 162 | + //累加前面的分割点,从j-minLength才能累加 |
| 163 | + if (canPartition(s, j - minLength)) sum = (sum + dp[i - 1][j - minLength]) % mod; |
| 164 | + //判断该点能不能分割 |
| 165 | + if (canPartition(s, j)) dp[i][j] = sum; |
| 166 | + } |
| 167 | + } |
| 168 | + return (int) dp[K][s.length()]; |
| 169 | + } |
| 170 | + |
| 171 | + private boolean isPrime(char c) { |
| 172 | + return c == '2' || c == '3' || c == '5' || c == '7'; |
| 173 | + } |
| 174 | + |
| 175 | + private boolean canPartition(String s, int j) { |
| 176 | + return j == 0 || j == s.length() || !isPrime(s.charAt(j - 1)) && isPrime(s.charAt(j)); |
| 177 | + } |
| 178 | + |
| 179 | + |
| 180 | + public static void main(String[] args) { |
| 181 | + } |
| 182 | +} |
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