Features Track 2018徐州icpc网络赛 思维

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_ixi​ = x_jxj​and y_iyi​ = y_jyj​, then <x_ixi​, y_iyi​> <x_jxj​, y_jyj​> 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_iki​ ( the number of features) and 2k_i2ki​intergers describe k_iki​ 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 NNwill 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

• • main.cpp

  

 

 

题意：n个集合，每个集合有ki个点，问连续在多个集合中出现的次数最多的点的次数

分析：用一个map存下每个点的最后一次出现位置和连续出现次数，连续出现次数通过最后一次出现位置来判断，每次更新连续出现次数时取下最大值

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e5+10;
const ll mod = 2e9+7;
const double pi = acos(-1.0);
const double eps = 1e-8;
map<ll,pair<ll,ll> >mp;
map<ll,ll> mm;
int main() {
    ll k, n, a, b, T;
    scanf("%lld",&T);
    while( T -- ) {
        scanf("%lld",&n);
        ll maxa = 0;
        mp.clear();
        for( ll i = 1; i <= n; i ++ ) {
            scanf("%lld",&k);
            mm.clear();
            while( k -- ) {
                scanf("%lld%lld",&a,&b);
                ll t = a*mod + b; //通过乘mod将每个坐标转化成一个不同的值
                if( mp[t].first == i-1 ) {
                    mp[t].second = mp[t].second + 1, mp[t].first = i;
                } else if( mm[t] ) {
                    continue;
                } else {
                    mp[t].second = 1, mp[t].first = i;
                }
                mm[t] ++;
                maxa = max(maxa,mp[t].second);
            }
        }
        if( maxa == 1 ) {
            maxa = 0;
        }
        printf("%lld\n",maxa);
    }
    return 0;
}

　　

转载于:https://www.cnblogs.com/l609929321/p/9614809.html