<CDQ分治> Jam's problem again [HDU - 5618]

最新推荐文章于 2021-04-01 22:36:43 发布

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最新推荐文章于 2021-04-01 22:36:43 发布

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原文链接：http://www.cnblogs.com/ikihsiguoyr/p/10569484.html

博客围绕HDU - 5618的三维偏序问题展开。该问题给定三维坐标轴上的点，根据点之间的大小关系确定点的层级。题解提出初始按x、y、z关键字排序，处理子问题时以y坐标归并排序，用树状数组维护前缀z信息，还给出相同点的处理方法。

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A - Jam's problem again HDU - 5618

Problem Description

Jam like to solve the problem which on the 3D-axis,given N(1≤N≤100000) points (x,y,z)(1≤x,y,z≤100000)

If two point such as$ (x_i,y_i,z_i)$ and $(x_j,y_j,z_j)$ $x_i≥x_j,y_i≥_j,z_i≥z_j$, the bigger one level add 1

Ask for the each level of the point.

Input

The first line is T(1≤T≤15) means T Case

For each case

The first line is N means the number of Point and next there are N line, each line has (x,y,z)

Output

Output with N line,each line has one number means the lever of point

Sample Input

Sample Output

题解

三维偏序

初始按$x$为意义关键字,$y$为第二关键字,$z$为第三关键字排序,一定不能省只按$x$排序,因为要处理相同的点

分别处理子问题,然后以$y$坐标为关键字, 将左右两端归并排序(这样递归左右内部$y$有序)

并用树状数组维护前缀的$z$的信息(当前左边的$x$一定比右边小,并且$y$也有序)

归并到左边的点就插入$z$, 右边的点询问$z$

对于相同的点,我们初始发现排完序后,相同的点中序号较前的点不会统计到后面相同的点,只有最后的那个点才统计完全

所以可以在一开始先将相同的点加上$p, p-1, ..., 1, 0$就好了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>

using namespace std;

typedef long long LL;
const int MAXN = 1e5 + 10;

inline LL in()
{
    LL x = 0, flag = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') flag = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
    return x * flag;
}

int T;

int n;

struct Fenwick
{
    int val[MAXN];
    void clear() { memset(val, 0, sizeof val); }
    void update(int x, int v)
        {
            for (int i = x; i <= 100000; i += i & (-i)) val[i] += v;
        }
    int query(int x)
        {
            int ret = 0;
            for (int i = x; i > 0; i &= (i - 1)) ret += val[i];
            return ret;
        }
} FT;

int an[MAXN];
struct Node
{
    int x, y, z, id;
    bool operator < (const Node & b) const
        {
            return x == b.x ? (y == b.y ? z < b.z : y < b.y) : x < b.x; 
        }
} a[MAXN], b[MAXN];

void merge(int l, int mid, int r)
{
    int i = l, j = mid + 1, k = i;
    while (k <= r)
    {
        if (j > r || (i <= mid && a[i].y <= a[j].y))
        {
            FT.update(a[i].z, 1);
            b[k ++] = a[i];
            ++ i;
        }
        else
        {
            an[a[j].id] += FT.query(a[j].z);
            b[k ++] = a[j];
            ++ j;
        }
    }
    for (int i = l; i <= mid; i ++) FT.update(a[i].z, -1);
    for (int i = l; i <= r; i ++) a[i] = b[i];
}
void solve(int l, int r)
{
    if (l >= r) return;
    int mid = (l + r) >> 1;
    solve(l, mid);
    solve(mid + 1, r);
    merge(l, mid, r);
}

int main()
{
    T = in();
    while (T --)
    {
        memset(a, 0, sizeof a);
        memset(an, 0, sizeof an);
        n = in(); 
        for (int i = 1; i <= n; i ++) a[i] = (Node) { in(), in(), in(), i };
        sort(a + 1, a + n + 1);
        for (int i = n - 1; i >= 1; i --)
        {
            if (a[i].x == a[i + 1].x && a[i].y == a[i + 1].y && a[i].z == a[i + 1].z) an[a[i].id] += an[a[i + 1].id] + 1;
        }
        solve(1, n);
        for (int i = 1; i <= n; i ++) printf("%d\n", an[i]);
    }
    return 0;
}


/*
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转载于:https://www.cnblogs.com/ikihsiguoyr/p/10569484.html