「CQOI2011」动态逆序对

「CQOI2011」动态逆序对

传送门
树套树。
删除一个位置的元素带来的减损数等于他前面大于它的和后面小于它的，然后这个直接树状数组套主席树维护一下就好了。
参考代码：

#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

typedef long long LL;
const int _ = 1e5 + 5, __ = 1e7 + 5;

int n, q, a[_], pos[_];
int tot, rt[_], lc[__], rc[__], t1[_], t2[_]; LL cnt[__];

inline void update(int& p, int x, int v, int l = 1, int r = n) {
    if (!p) p = ++tot; cnt[p] += v;
    if (l == r) return ;
    int mid = (l + r) >> 1;
    if (x <= mid) update(lc[p], x, v, l, mid);
    else update(rc[p], x, v, mid + 1, r);
}

inline LL Query(int l, int r, int x, int opt) {
    int c1 = 0, c2 = 0; --l;
    for (rg int i = l; i >= 1; i -= i & -i) t1[++c1] = rt[i];
    for (rg int i = r; i >= 1; i -= i & -i) t2[++c2] = rt[i];
    l = 1, r = n;
    LL res = 0;
    while (l < r) {
        int mid = (l + r) >> 1;
        if (x <= mid) {
            if (opt == 0) {
                for (rg int i = 1; i <= c1; ++i) res -= cnt[rc[t1[i]]];
                for (rg int i = 1; i <= c2; ++i) res += cnt[rc[t2[i]]];
            }
            for (rg int i = 1; i <= c1; ++i) t1[i] = lc[t1[i]];
            for (rg int i = 1; i <= c2; ++i) t2[i] = lc[t2[i]];
            r = mid;
        } else {
            if (opt == 1) {
                for (rg int i = 1; i <= c1; ++i) res -= cnt[lc[t1[i]]];
                for (rg int i = 1; i <= c2; ++i) res += cnt[lc[t2[i]]];             
            }
            for (rg int i = 1; i <= c1; ++i) t1[i] = rc[t1[i]];
            for (rg int i = 1; i <= c2; ++i) t2[i] = rc[t2[i]];         
            l = mid + 1;
        }
    }
    return res;
}

int main() {
    read(n), read(q);
    LL ans = 0;
    for (rg int i = 1; i <= n; ++i) {
        read(a[i]), pos[a[i]] = i;
        ans += Query(1, i - 1, a[i], 0);
        for (rg int j = i; j <= n; j += j & -j) update(rt[j], a[i], 1);
    }
    for (rg int x; q--; ) {
        printf("%lld\n", ans);
        read(x);
        ans -= Query(1, pos[x] - 1, x, 0);
        ans -= Query(pos[x] + 1, n, x, 1);
        for (rg int j = pos[x]; j <= n; j += j & -j) update(rt[j], x, -1);
    }
    return 0;
}