2018徐州网络预赛 F题：Features Track

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本文探讨了一个计算机视觉问题，即从猫的视频中识别猫的运动轨迹。通过对每一帧的猫特征进行提取，并将这些特征视为二维向量，文章提出了一种算法来寻找最长的连续猫特征序列，以此判断猫的运动。输入包含多个测试用例，每个用例有多帧图像，每帧可能包含不同数量的猫特征。输出则是最长的猫特征运动序列长度。

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题目链接：https://nanti.jisuanke.com/t/A2005

Morgana is learning computer vision, and he likes cats, too. One day he wants to find the cat movement from a cat video. To do this, he extracts cat features in each frame. A cat feature is a two-dimension vector <xx, yy>. If x_ix
i

= x_jx
j

and y_iy
i

= y_jy
j

, then <x_ix
i

, y_iy
i

<x_jx
j

, y_jy
j

are same features.

So if cat features are moving, we can think the cat is moving. If feature <aa, bb> is appeared in continuous frames, it will form features movement. For example, feature <aa , bb > is appeared in frame 2,3,4,7,82,3,4,7,8, then it forms two features movement 2-3-42−3−4 and 7-87−8 .

Now given the features in each frames, the number of features may be different, Morgana wants to find the longest features movement.

Input
First line contains one integer T(1 \le T \le 10)T(1≤T≤10) , giving the test cases.

Then the first line of each cases contains one integer nn (number of frames),

In The next nn lines, each line contains one integer k_ik
i

( the number of features) and 2k_i2k
i

intergers describe k_ik
i

features in ith frame.(The first two integers describe the first feature, the 33rd and 44th integer describe the second feature, and so on).

In each test case the sum number of features NN will satisfy N \le 100000N≤100000 .

Output
For each cases, output one line with one integers represents the longest length of features movement.

样例输入复制
1
8
2 1 1 2 2
2 1 1 1 4
2 1 1 2 2
2 2 2 1 4
0
0
1 1 1
1 1 1
样例输出复制
3
题目来源
ACM-ICPC 2018 徐州赛区网络预赛

分析：

假如是有间接的容器存次数的话，要注意下面这组数据

代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <map>
#include<set>

using namespace std;
const int N=1e5+5;
#define p pair<int,int>


map<p,int> mp1;
map<int,int> mp2;

set<int> st[2];


int main()
{
	int t;
	int n;
	int k;
	int x,y;
	scanf("%d",&t);
	while(t--)
	{
		mp1.clear();
		mp2.clear();
		scanf("%d",&n);
		int mx=0;
		int cnt=1;
		int cur=1;
		st[0].clear();
		st[1].clear();
		for(int i=0;i<n;i++)
		{
			scanf("%d",&k);
			cur^=1;	
			st[cur].clear();
			for(int j=0;j<k;j++)
			{
				scanf("%d %d",&x,&y);
				p w=make_pair(x,y);
				if(mp1[w]<1)
				{
					mp1[w]=cnt++;
				}
				int ps=mp1[w];
				
				if(st[cur^1].count(ps))
				{
					if(!st[cur].count(ps))
						mp2[ps]+=1;
				}
				else
					mp2[ps]=1;
				st[cur].insert(ps);
			
				mx=max(mx,mp2[ps]);				
				
			}
			
			
		}
		cout<<mx<<endl;
	}
	return 0;
}