2018 Nowcoder Multi-University Training Contest 1

Practice Link

J. Different Integers

題意：
給出\(n\)個數，每次詢問\((l_i, r_i)\)，表示\(a_1, \cdots, a_i, a_j, \cdots, a_n\)中有多少個不一樣的數。

思路：
先分別離線求出\(a_1, \cdots a_i\)以及\(a_j, \cdots, a_n\)中有多少個不一樣的數。
再考慮有多少個數既在\([1, i]\)中也在\([j, n]\)中，再離線作一次。
考慮一個數第一次出現的時候，那麼這個數下一次出現的位置以及以後的全部詢問區間都要減去一個貢獻。

代碼：

#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;
}