【AtCoder Beginner Contest 330)】[E - Mex and Update ] 线段树+二分

本题可以用线段树+二分的方式实现。代码如下：

import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;

// Press Shift twice to open the Search Everywhere dialog and type `show whitespaces`,
// then press Enter. You can now see whitespace characters in your code.
public class Main {
    public static void main(String[] args) throws IOException {
        int n, q;
        StreamTokenizer st = new StreamTokenizer(new InputStreamReader(System.in));
        st.nextToken();
        n = (int) st.nval;
        st.nextToken();
        q = (int) st.nval;
        int[] a = new int[n];
        tree = new int[4 * n];
        int[] b = new int[n + 1];
        int[] cnt = new int[n + 1];
        for (int i = 0; i < n; i++) {
            st.nextToken();
            a[i] = (int) st.nval;
            if (a[i] <= n) {
                cnt[a[i]]++;
                if (b[a[i]] == 0) {
                    b[a[i]]++;
                }
            }
        }
        PrintWriter writer = new PrintWriter(System.out);
        build(0, b, 0, n);
        for (int i = 0; i < q; i++) {
            st.nextToken();
            int index = (int) st.nval-1;
            st.nextToken();
            int value = (int) st.nval;

            int oldvalue = a[index];

            if (oldvalue <= n) {
                cnt[oldvalue]--;
                if (cnt[oldvalue] == 0) {
                    b[oldvalue]--;
                    add(oldvalue, -1, 0, 0, n);
                }
            }

            a[index] = value;
            if (value <= n) {
                cnt[value]++;
                if (b[value] == 0) {
                    b[value]++;
                    add(value, 1, 0, 0, n);
                }
            }


            //二分
            int left = 0, right = n;
            while (left < right) {
                int mid = (left + right) / 2;
                if(query(0,mid,0,0,n) == mid+1){
                    left = mid + 1;
                }else{
                    right = mid;
                }
            }
//            System.out.println(left);
            writer.println(left);
        }
        writer.flush();
    }

    static int[] tree;

    public static int query(int l, int r, int index,int L, int R) {
        int ans = 0;
        if (l <= L && R <= r) {
            return tree[index];
        }else {
            int mid = (L + R) / 2;
            if(mid<l){
                return query(l,r,2*index+2,mid+1,R);
            }
            else if(r<=mid){
                return query(l,r,2*index+1,L,mid);
            }else{
                return query(l,mid,2*index+1,L,mid)+query(mid+1,r,2*index+2,mid+1,R);
            }
        }
    }

    public static void build(int index, int[] arr, int l, int r) {
        if (l == r) {
            tree[index] = arr[l];
            return;
        }
        build(2 * index + 1, arr, l, (l + r) / 2);
        build(2 * index + 2, arr, (l + r) / 2 + 1, r);
        tree[index] = tree[2 * index + 1] + tree[2 * index + 2];
    }

    public static void add(int l, int v, int index, int L, int R) {
        if (L == R) {
            tree[index] += v;
            return;
        }
        int mid = (L + R) / 2;
        if (l <= mid) {
            add(l, v, 2 * index + 1, L, mid);
        } else {
            add(l, v, 2 * index + 2, mid + 1, R);
        }
        tree[index] = tree[2 * index + 1] + tree[2 * index + 2];
    }


}