2018 Nowcoder Multi-University Training Contest 1

Practice Link

J. Different Integers

题意：
给出\(n\)个数，每次询问\((l_i, r_i)\)，表示\(a_1, \cdots, a_i, a_j, \cdots, a_n\)中有多少个不同的数。

思路：
先分别离线求出\(a_1, \cdots a_i\)以及\(a_j, \cdots, a_n\)中有多少个不同的数。
再考虑有多少个数既在\([1, i]\)中也在\([j, n]\)中，再离线做一次。
考虑一个数第一次出现的时候，那么这个数下一次出现的位置以及之后的所有询问区间都要减去一个贡献。

代码：

#include <bits/stdc++.h>
using namespace std;

#define N 100010
int n, q, a[N], b[N], c[N], nx[N], ans[N];
struct node {
    int l, r, id;
    node() {}
    void scan(int id) {
        this->id = id;
        scanf("%d%d", &l, &r);
        if (l >= r) {
            l = 1;
            r = 2;
        }
    }
}qrr[N];

struct BIT {
    int a[N];
    void init() {
        memset(a, 0, sizeof a);
    }
    void update(int x, int v) {
        for (; x > 0; x -= x & -x) {
            a[x] += v;
        }
    }
    int query(int x) {
        int res = 0;
        for (; x < N; x += x & -x) {
            res += a[x];
        }
        return res;
    }
    int query(int l, int r) {
        return query(l) - query(r + 1);
    }
}bit; 

int main() {
    while (scanf("%d%d", &n, &q) != EOF) {
        for (int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
        }
        for (int i = 1; i <= q; ++i) {
            qrr[i].scan(i);
        }
        if (n == 1) {
            for (int i = 1; i <= q; ++i) {
                printf("1\n");
            }
            continue;
        }
        sort(qrr + 1, qrr + 1 + q, [&](node x, node y) {
            return x.l < y.l;
        });
        memset(b, 0, sizeof b);
        for (int i = 1, j = 1, k = 0; i <= q; ++i) {
            while (j <= n && j <= qrr[i].l) {
                if (b[a[j]] == 0) {
                    b[a[j]] = 1;
                    ++k;
                }
                ++j;
            }
            ans[qrr[i].id] = k;
        }
        sort(qrr + 1, qrr + 1 + q, [&](node x, node y){
            return x.r > y.r;
        });
        memset(b, 0, sizeof b);
        for (int i = 1, j = n, k = 0; i <= q; ++i) {
            while (j >= 1 && j >= qrr[i].r) {
                if (b[a[j]] == 0) {
                    b[a[j]] = 1;
                    ++k;
                }
                --j;
            }
            ans[qrr[i].id] += k;
        }
        memset(b, 0, sizeof b);
        for (int i = 1; i <= n; ++i) {
            nx[i] = n + 1;
        }
        for (int i = n; i >= 1; --i) {
            c[i] = nx[a[i]];
            if (nx[a[i]] == n + 1) {
                nx[a[i]] = i;
            }
        }
        bit.init();
        sort(qrr + 1, qrr + 1 + q, [&](node x, node y){
            return x.l < y.l;
        });
        for (int i = 1, j = 1; i <= q; ++i) {
            while (j <= n && j <= qrr[i].l) {
                if (b[a[j]] == 0) {
                    bit.update(c[j], -1);
                    b[a[j]] = 1;
                }
                ++j;
            }
            ans[qrr[i].id] += bit.query(qrr[i].r, n);
        }
        for (int i = 1; i <= q; ++i) {
            printf("%d\n", ans[i]);
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/Dup4/p/11108577.html