[洛谷P2709]小B的询问

题目传送门

这道题的解法是莫队。莫队是一个离线的算法（也可以在线），能满足$O(1)$转移的算法。通过分块来获得适当的求解顺序使得复杂度控制在$O(nsqrt{n})$。

#include <bits/stdc++.h>

using namespace std;

#define re register
#define rep(i, a, b) for (re int i = a; i <= b; ++i)
#define repd(i, a, b) for (re int i = a; i >= b; --i)
#define maxx(a, b) a = max(a, b);
#define minn(a, b) a = min(a, b);
#define LL long long
#define INF (1 << 30)

inline int read() {
    int w = 0, f = 1; char c = getchar();
    while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar();
    while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar();
    return w * f;
}

const int maxn = 50000 + 5;

struct Block {
    int l, r, pos, id, ans;
} q[maxn];
bool cmp1(Block a, Block b) { return a.pos < b.pos || a.pos == b.pos && a.r < b.r; }
bool cmp2(Block a, Block b) { return a.id < b.id; }
int sqr(int a) { return a*a; }

int n, m, k, a[maxn], size, cnt[maxn], ans = 0;

int main() {
    n = read(), m = read(), k = read();
    size = sqrt(n);
    rep(i, 1, n)
        a[i] = read();

    rep(i, 1, m) q[i].l = read(), q[i].r = read(), q[i].id = i, q[i].pos = (q[i].l+size-1) / size;
    sort(q+1, q+m+1, cmp1);

    int pl = 1, pr = 0;
    rep(i, 1, m) {
        while (pl < q[i].l) ans -= 2 * cnt[a[pl]] - 1, cnt[a[pl++]]--; 
        while (pl > q[i].l) cnt[a[--pl]]++, ans += 2 * cnt[a[pl]] - 1;
        while (pr < q[i].r) cnt[a[++pr]]++, ans += 2 * cnt[a[pr]] - 1;
        while (pr > q[i].r) ans -= 2 * cnt[a[pr]] - 1, cnt[a[pr--]]--;
        q[i].ans = ans;
    }
    sort(q+1, q+m+1, cmp2);
    rep(i, 1, m) printf("%d
", q[i].ans);

    return 0;
}