hdu 5517 Triple (二维树状数组)

最新推荐文章于 2021-05-23 19:02:44 发布

Timeclimber

最新推荐文章于 2021-05-23 19:02:44 发布

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分类专栏：树状数组思维文章标签： hdu 二维树状数组

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本文介绍了一个算法问题，涉及多集合的定义与运算，通过构造特定的三元组并利用离线二维树状数组来解决一个关于TOP子集大小计算的问题。

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Triple

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1127 Accepted Submission(s): 403

Problem Description

Given the finite multi-set

A of

n pairs of integers, an another finite multi-set

B of

m triples of integers, we define the product of

A and

B as a multi-set

C=A∗B={⟨a,c,d⟩∣⟨a,b⟩∈A, ⟨c,d,e⟩∈B and b=e}

For each

⟨a,b,c⟩∈C , its BETTER set is defined as

BETTERC(⟨a,b,c⟩)={⟨u,v,w⟩∈C∣⟨u,v,w⟩≠⟨a,b,c⟩, u≥a, v≥b, w≥c}

As a \textbf{multi-set} of triples, we define the TOP subset (as a multi-set as well) of

C , denoted by

TOP(C) , as

TOP(C)={⟨a,b,c⟩∈C∣BETTERC(⟨a,b,c⟩)=∅}

You need to compute the size of

TOP(C) .

Input

The input contains several test cases. The first line of the input is a single integer

t (1≤t≤10) which is the number of test case. Then

t test cases follow.

Each test case contains three lines. The first line contains two integers

n (1≤n≤105) and

m (1≤m≤105) corresponding to the size of

A and

B respectively.
The second line contains

2×n nonnegative integers

a 1, b 1, a 2, b 2, \dots, a n, b n

which describe the multi-set

A , where

1≤ai,bi≤105 .
The third line contains

3×m nonnegative integers

c 1, d 1, e 1, c 2, d 2, e 3, \dots, c m, d m, e m

corresponding to the

m triples of integers in

B , where

1≤ci,di≤103 and

1≤ei≤105 .

Output

For each test case, you should output the size of set

TOP(C) .

Sample Input

  
   2
5 9
1 1 2 2 3 3 3 3 4 2
1 4 1 2 2 1 4 1 1 1 3 2 3 2 2 4 1 2 2 4 3 3 2 3 4 1 3
3 4
2 7 2 7 2 7
1 4 7 2 3 7 3 2 7 4 1 7

Sample Output

  
   Case #1: 5
Case #2: 12

Source

2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

Recommend

wange2014

题意：

给了n个二元集<a,b>和m个三元集<c,d,e>。给出了构造另一个元集合的方法，当b==e时构造成三元集C={<a,c,d>}。求top(C)集合。

思路：

第一：当e相同的时候，若可构造多个集合<a1,b,c>,<a2,b,c>...<an,b,c>则取<max(a1,a2,...an)>这个集合。因为小于最大值的都会被舍弃。

第二：去除它的重点。用一个cnt来记录它的重点的个数。为了方便后面的计算。

第三：可以将a,b,c依次按降序来排，然后从后往前面遍历。由b和c构成一个二维平面，每次加点之前看它的右上区域是否还有点，如果有点则该点有Better集合，所以不能加。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[100005];
int cnt[100005];
int tr[1005][1005];
int lowbit(int i)
{
    return i&(-i);
}
void update(int x,int y,int n)
{
    for(int i=x;i<=n;i+=lowbit(i))
    {
        for(int j=y;j<=n;j+=lowbit(j))
        {
            tr[i][j]++;
        }
    }
}
int query(int x,int y)
{
    int sum=0;
    for(int i=x;i>0;i-=lowbit(i))
    {
        for(int j=y;j>0;j-=lowbit(j))
        {
            sum+=tr[i][j];
        }
    }
    return sum;
}
struct node
{
    int a,b,c,cnt;
    node()
    {
        cnt=0;
    }
}q[100005];
bool cmp(node a,node b)
{
    if(a.a!=b.a)
        return a.a<b.a;
    else if(a.b!=b.b)
        return a.b<b.b;
    else
        return a.c<b.c;
}
int main()
{
    int t,n,m,x,y,z,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        memset(cnt,0,sizeof(cnt));
        memset(tr,0,sizeof(tr));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&x,&y);
            if(a[y]==0)
            {
                a[y]=x;
                cnt[x]=1;
            }
            else
            {
                if(x>a[y])
                {
                    a[y]=x;
                    cnt[x]=1;
                }
                else if(x==a[y])
                    cnt[x]++;
            }
        }
        int p=0;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            if(cnt[a[z]]==0)
                continue;
            else
            {
                q[p].a=a[z];
                q[p].b=x;
                q[p].c=y;
                q[p].cnt=cnt[a[z]];
                p++;
            }
        }
        sort(q,q+p,cmp);
        int k=0;
        for(int i=1;i<p;i++)
        {
            if(q[k].a==q[i].a&&q[k].b==q[i].b&&q[k].c==q[i].c)
            {
                q[k].cnt+=q[i].cnt;
            }
            else
            {
                q[++k]=q[i];
            }
        }
        int ans=0;
        for(int i=k;i>=0;i--)
        {
            int num=query(1001,1001)-query(1001,q[i].c-1)-query(q[i].b-1,1001)+query(q[i].b-1,q[i].c-1);
            if(num==0) ans+=q[i].cnt;
            update(q[i].b,q[i].c,1001);
        }
        printf("Case #%d: %d\n",cas++,ans);
    }
    return 0;
}