• [洛谷P3810]【模板】三维偏序(陌上花开)


    题目大意:有$n$个元素,第$i$个元素有三个属性$a_i,b_i,c_i$,设$f(i)=sumlimits_{i ot = j}[a_jleqslant a_i,b_jleqslant b_i,c_jleqslant c_i]$,令$d(i)=sumlimits_{j=1}^n[f(j)=i]$,求$d$

    题解:三位偏序,我用了$CDQ$分治,$a$排序解决,$b$$CDQ$分治,$c$用树状数组

    卡点:

    C++ Code:

    #include <cstdio>
    #include <algorithm>
    #define maxn 100010
    #define maxm 200010
    int n, k, tot;
    int d[maxn];
    struct node {
        int a, b, c, cnt, f;
        inline bool operator == (const node &rhs) const {
            return (a == rhs.a && b == rhs.b && c == rhs.c);
        }
    } v[maxn], s[maxn];
    inline bool cmpb(node lhs, node rhs) {return lhs.b == rhs.b ? lhs.c < rhs.c : lhs.b < rhs.b;}
    inline bool cmpa(node lhs, node rhs) {return lhs.a == rhs.a ? cmpb(lhs, rhs) : lhs.a < rhs.a;}
    namespace Binary_Indexed_Tree {
        int Tr[maxm];
        int res;
        inline void add(int p, int num) {for (; p <= k; p += p & -p) Tr[p] += num;}
        inline int ask(int p) {res = 0; for (; p; p &= p - 1) res += Tr[p]; return res;}
    }
    using Binary_Indexed_Tree::add;
    using Binary_Indexed_Tree::ask;
    void CDQ(int l, int r) {
        if (l == r) return ;
        int mid = l + r >> 1;
        CDQ(l, mid); CDQ(mid + 1, r);
        std::sort(s + l, s + mid + 1, cmpb);
        std::sort(s + mid + 1, s + r + 1, cmpb);
        int pl = l, pr = mid + 1;
        while (pl <= mid && pr <= r) {
            if (s[pl].b <= s[pr].b) add(s[pl].c, s[pl].cnt), pl++;
            else s[pr].f += ask(s[pr].c), pr++;
        }
        while (pr <= r) s[pr].f += ask(s[pr].c), pr++;
        for (int i = l; i < pl; i++) add(s[i].c, -s[i].cnt);
    }
    int main() {
        scanf("%d%d", &n, &k);
        for (int i = 1; i <= n; i++) scanf("%d%d%d", &v[i].a, &v[i].b, &v[i].c);
        std::sort(v + 1, v + n + 1, cmpa);
        for (int i = 1, j; (j = i) <= n; i = j) {
            while (j <= n && v[i] == v[j]) j++;
            s[++tot] = v[i]; s[tot].cnt = j - i;
        }
        CDQ(1, tot);
        for (int i = 1; i <= tot; i++) d[s[i].f + s[i].cnt] += s[i].cnt;
        for (int i = 1; i <= n; i++) printf("%d
    ", d[i]);
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Memory-of-winter/p/9701110.html
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