洛谷 P1972 [SDOI2009] HH的项链

题目链接

https://www.luogu.org/problem/P1972

分析

将询问离线，同时要用到树状数组。
先把询问按右端点升序排序，依次处理，此时对于重复出现的数，我们实际上只关心当前最靠右的。

AC代码

#include 
#include 

using namespace std;

inline int get_num() {
    int num = 0;
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    while (c >= '0' && c <= '9')
        num = num * 10 + c - '0', c = getchar();
    return num;
}

const int maxn = 5e5 + 5, maxc = 1e6 + 5;

struct Question {
    int l, r, id;
    bool operator < (const Question& rhs) const {
        return r < rhs.r;
    }
} q[maxn];

int n, m, num[maxn], last[maxc], bit[maxn], ans[maxn];

inline int ask(int x) {
    int sum = 0;
    while (x) sum += bit[x], x -= x & (-x);
    return sum;
}

inline void add(int x, int d) {
    while (x <= n)
        bit[x] += d, x += x & (-x);
}

int main() {
    n = get_num();
    for (int i = 1; i <= n; ++i) num[i] = get_num();
    m = get_num();
    for (int i = 1; i <= m; ++i)
        q[i].l = get_num(), q[i].r = get_num(), q[i].id = i;
    sort(q + 1, q + m + 1);
    int j = 0;
    for (int i = 1; i <= m; ++i) {
        while (j < q[i].r) {
            ++j;
            if (last[num[j]]) add(last[num[j]], -1);
            add(j, 1);
            last[num[j]] = j;
        }
        ans[q[i].id] = ask(q[i].r) - ask(q[i].l - 1);
    }
    for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
    return 0;
}