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  • BZOJ 3289 Mato的文件管理

    莫队算法+树状数组转移

    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<queue>
    #include<algorithm>
    using namespace std;
    
    const int maxn = 50000 + 10;
    int n, m;
    int tmp[maxn], a[maxn], lsh[maxn], cnt;
    int pos[maxn];
    int c[maxn];
    long long ans[maxn], Ans;
    int L, R;
    
    struct X
    {
        int l, r, id;
    }s[maxn];
    
    int f(int x)
    {
        int l = 1, r = cnt;
        while (l <= r)
        {
            int mid = (l + r) / 2;
            if (lsh[mid] < x) l = mid + 1;
            else if (lsh[mid]>x) r = mid - 1;
            else return mid;
        }
    }
    
    void LSH()
    {
        sort(tmp + 1, tmp + 1 + n);
        cnt = 0, lsh[++cnt] = tmp[1];
        for (int i = 2; i <= n; i++)
        {
            if (tmp[i] == tmp[i - 1]) continue;
            lsh[++cnt] = tmp[i];
        }
        for (int i = 1; i <= n; i++) a[i] = f(a[i]);
    }
    
    bool cmp(const X&a, const X&b)
    {
        if (pos[a.l] == pos[b.l]) return a.r < b.r;
        return a.l < b.l;
    }
    
    int lowbit(int x) { return x&(-x); }
    void update(int pos, int val)
    {
        while (pos <= cnt) 
            c[pos] = c[pos] + val, pos = pos + lowbit(pos);
    }
    int getsum(int pos)
    {
        int res = 0;
        while (pos > 0) 
            res = res + c[pos], pos = pos - lowbit(pos); 
        return res;
    }
    
    int main()
    {
        while (~scanf("%d", &n)) {
            int sz = sqrt(n);
            for (int i = 1; i <= n; i++)
            {
                pos[i] = i / sz;
                scanf("%d", &tmp[i]);
                a[i] = tmp[i];
            }
            LSH();
    
            scanf("%d", &m);
            for (int i = 1; i <= m; i++) {
                scanf("%d%d", &s[i].l, &s[i].r);
                s[i].id = i;
            }
            sort(s + 1, s + 1 + m, cmp);
    
            for (int i = s[1].l; i <= s[1].r; i++)
            {
                int sum = getsum(cnt) - getsum(a[i]);
                Ans = Ans + (long long)sum;
                update(a[i], 1);
            }
            ans[s[1].id] = Ans; L = s[1].l; R = s[1].r;
    
            for (int i = 2; i <= m; i++)
            {
                while (L < s[i].l)
                {
                    int sum = getsum(a[L] - 1);
                    Ans = Ans - (long long)sum;
                    update(a[L], -1);
                    L++;
                }
                while (L > s[i].l)
                {
                    L--;
                    int sum = getsum(a[L] - 1);
                    Ans = Ans + (long long)sum;
                    update(a[L], 1);
                }
                while (R < s[i].r)
                {
                    R++;
                    int sum = getsum(cnt) - getsum(a[R]);
                    Ans = Ans + (long long)sum;
                    update(a[R], 1);
                }
                while (R > s[i].r)
                {
                    int sum = getsum(cnt) - getsum(a[R]);
                    Ans = Ans - (long long)sum;
                    update(a[R], -1);
                    R--;
                }
                ans[s[i].id] = Ans;
            }
    
            for (int i = 1; i <= m; i++) printf("%lld
    ", ans[i]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zufezzt/p/5412210.html
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