Codeforces Round 926 (Div. 2) ---- E. Sasha and the Happy Tree Cutting ----题解

E. Sasha and the Happy Tree Cutting：

题目大意：

思路解析：

现在有一颗树，然后给出了k对路径，然后要求路径上至少有一个结点是被染色了的，如果这k对路径没有共用边，那我们至少需要染色k条边。如果有共用边，那么选择将他染色，那么就可以使多个路径满足条件。现在求最小是染色树。已知最多需要20条边，那我们可以找出哪些边可以进行染色（并且它染色后能解决哪些路径），然后进行状压dp。

代码实现：

import java.util.*;


import java.io.*;
import java.math.*;

public class Main {
    static int[] pre;
    static Vector<int[]>[] g;
    public static void main(String[] args) throws IOException {
        int t = f.nextInt();
        while (t > 0) {
            solve();
            t--;
        }
        w.flush();
        w.close();
    }

    public static void solve() throws IOException{
        int n = f.nextInt();
        g = new Vector[n+1];
        pre = new int[n+1];
        for (int i = 0; i < n + 1; i++) {
            g[i] = new Vector<>();
        }
        for (int i = 0; i < n - 1; i++) {
            int x = f.nextInt();
            int y = f.nextInt();
            g[x].add(new int[] {y, i});
            g[y].add(new int[] {x, i});
        }
        int[] st = new int[n];
        dfs1(1, 0);
        int k = f.nextInt();
        for (int i = 0; i < k; i++) {
            int x = f.nextInt(); int y = f.nextInt();
            if (pre[x] != 0) dfs2(x);

            while ((x ^ y) != 0){
                for (int j = 0; j < g[y].size(); j++) {
                    int[] cur = g[y].get(j);
                    if (cur[0] == pre[y]){
                        y = pre[y];
                        st[cur[1]] |= (1 << i);
                        break;
                    }
                }
            }
        }
        HashSet<Integer> set = new HashSet<>();
        for (int i = 0; i < n; i++) {
            if (st[i] > 0)
                set.add(st[i]);
        }
        int[] dp = new int[1 << k];
        Arrays.fill(dp, 1000);
        dp[0] = 0;
        for (int i = 0; i < (1 << k); i++) {
            for (Integer j: set){
                dp[i | j] = Math.min(dp[i|j], dp[i] + 1);
            }
        }
        w.println(dp[(1 << k) - 1]);
    }

    public static void dfs1(int x, int fa){
        for (int i = 0; i < g[x].size(); i++) {
            int[] cur = g[x].get(i);
            int y = cur[0];
            if (y == fa) continue;
            pre[y] = x;
            dfs1(y, x);
        }

    }

    public static void dfs2(int x){
        int y = pre[x];
        pre[x] = 0;
        while (y != 0){
            int t = pre[y];
            pre[y] = x;
            x = y;
            y = t;
        }
    }



    static PrintWriter w = new PrintWriter(new OutputStreamWriter(System.out));
    static Input f = new Input(System.in);

    static class Input {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public Input(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() throws IOException{
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(reader.readLine());
            }
            return tokenizer.nextToken();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public Double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
    }
}