Features Track
31.32% 1000ms 262144KMorgana 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.
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_i2kiintergers 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 NN will satisfy N \le 100000N≤100000 .
For each cases, output one line with one integers represents the longest length of features movement.
ACM-ICPC 2018 徐州赛区网络预赛
二维map记录当前行累积pair个数，因为只涉及到当前行与上一行的状态，因此采用01滚动map即可。
#include<bits/stdc++.h> using namespace std; typedef long long ll; map<int,map<pair<int,int>,int> > mp; pair<int,int> p; int main() { int t,n,x,y,i,j,k; scanf("%d",&t); while(t--){ scanf("%d",&n); int ans=0,xxx=0; for(int i=0;i<n;i++){ scanf("%d",&k); mp[xxx].clear(); while(k--){ scanf("%d%d",&x,&y); p = make_pair(x,y); if(mp[xxx^1].count(p)){ mp[xxx][p] = mp[xxx^1][p]+1; ans=max(ans,mp[xxx][p]); } else{ mp[xxx][p]=1; ans=max(ans,mp[xxx][p]); } } xxx^=1; } printf("%d\n",ans); } return 0; }
转载于:https://www.cnblogs.com/yzm10/p/9614228.html
