[Bzoj3262]陌上花开(CDQ分治&&树状数组||树套树)

题目链接

题目就是赤裸裸的三维偏序，所以用CDQ+树状数组可以比较轻松的解决，但是还是树套树好想QAQ

CDQ+树状数组

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll maxn =  + ;
struct node {
    int a, b, c, num, val;
}q[maxn], w[maxn];
bool cmp(node x, node y) {
    return x.a == y.a ? (x.b == y.b ? x.c < y.c : x.b < y.b) : x.a < y.a;
}
int sum[maxn], ans[maxn];
int n, k;
int lowbit(int x) {
    return x & -x;
}
void add(int x, int val) {
    while (x <= k) {
        sum[x] += val;
        x += lowbit(x);
    }
}
int query(int x) {
    int ans = ;
    while (x > ) {
        ans += sum[x];
        x -= lowbit(x);
    }
    return ans;
}
void CDQ(int l, int r) {
    if (l == r)return;
    int mid = l + r >> ;
    CDQ(l, mid); CDQ(mid + , r);
    int L = l, R = mid + , cnt = l;
    while (L <= mid && R <= r) {
        if (w[L].b <= w[R].b)add(w[L].c, w[L].num), q[cnt++] = w[L++];
        else w[R].val += query(w[R].c), q[cnt++] = w[R++];
    }
    while (L <= mid)add(w[L].c, w[L].num), q[cnt++] = w[L++];
    while (R <= r)w[R].val += query(w[R].c), q[cnt++] = w[R++];
    for (int i = l; i <= mid; i++)add(w[i].c, -w[i].num);
    for (int i = l; i <= r; i++)w[i] = q[i];
}
int main() {
    scanf("%d%d", &n, &k);
    for (int i = ; i <= n; i++)
        scanf("%d%d%d", &q[i].a, &q[i].b, &q[i].c), q[i].num = ;
    sort(q + , q +  + n, cmp);
    int cnt = ;
    w[] = q[];
    for (int i = ; i <= n; i++) {
        if (q[i].a == w[cnt].a&&q[i].b == w[cnt].b&&q[i].c == w[cnt].c)w[cnt].num++;
        else w[++cnt] = q[i];
    }
    CDQ(, cnt);
    for (int i = ; i <= cnt; i++)
        ans[w[i].val + w[i].num - ] += w[i].num;
    for (int i = ; i < n; i++)
        printf("%d\n", ans[i]);
}

树套树（树状数组套线段树）

因为空间有限，线段树要动态开点且要写成链表QAQ。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned long long ull;
 const ll maxn =  + ;
 struct node {
     int a, b, c, num, val;
 }q[maxn], w[maxn];
 bool cmp(node x, node y) {
     return x.a == y.a ? (x.b == y.b ? x.c < y.c : x.b < y.b) : x.a < y.a;
 }
 int n, k;
 int ans[maxn];
 struct seg {
     struct node {
         int  val;
         node *ls, *rs;
         node() { val = ; ls = rs = NULL; }
     };
     node *root;
     seg() {
         root = NULL;
     }
     void update(node *&x, int pos, int val, int l, int r) {
         if (!x)x = new node();
         if (l == r) {
             x->val += val;
             return;
         }
         int mid = l + r >> ;
         if (pos <= mid)update(x->ls, pos, val, l, mid);
         else update(x->rs, pos, val, mid + , r);
         x->val = (x->ls ? x->ls->val : ) + (x->rs ? x->rs->val : );
     }
     int query(node *x, int L, int R, int l, int r) {
         if (!x)return ;
         if (L <= l && r <= R)
             return x->val;
         int mid = l + r >> , ans = ;
         if (L <= mid)ans += query(x->ls, L, R, l, mid);
         if (R > mid)ans += query(x->rs, L, R, mid + , r);
         return ans;
     }
 }T[maxn];
 int lowbit(int x) {
     return x & -x;
 }
 void add(int x, int q, int val) {
     while (x <= k) {
         T[x].update(T[x].root, q, val, , k);
         x += lowbit(x);
     }
 }
 int query(int x, int q) {
     int ans = ;
     while (x > ) {
         ans += T[x].query(T[x].root, , q, , k);
         x -= lowbit(x);
     }
     return ans;
 }
 int main() {
     scanf("%d%d", &n, &k);
     for (int i = ; i <= n; i++)
         scanf("%d%d%d", &q[i].a, &q[i].b, &q[i].c), q[i].num = ;
     sort(q + , q +  + n, cmp);
     int cnt = ;
     w[] = q[];
     for (int i = ; i <= n; i++) {
         if (q[i].a == w[cnt].a&&q[i].b == w[cnt].b&&q[i].c == w[cnt].c)w[cnt].num++;
         else w[++cnt] = q[i];
     }
     for (int i = ; i <= cnt; i++) {
         add(w[i].b, w[i].c, w[i].num);
         int p = query(w[i].b, w[i].c);
         ans[p] += w[i].num;
     }
     for (int i = ; i <= n; i++)
         printf("%d\n", ans[i]);
 }