# PinkAndBlue

ptrdung

plain_text

2 months ago

5.0 kB

0

Indexable

Never

^{}

Pink and Blue Xenny was a teacher and he had N students. The N children were sitting in a room. Each child was wearing a white T-shirt, with a unique number from the range 1 to N written on it. T-Shirts of pink and blue color were to be distributed among the students by Xenny. This made the students very happy. Xenny felt that a random distribution of T-Shirts would be very uninteresting. So, he decided to keep an interesting condition: Every student would get a T-Shirt that is of a different color than his/her friends. That is, if X and Y are friends and X has a Pink T-Shirt, then Y should compulsorily have a Blue T-Shirt, and vice-versa. Also, Xenny had a belief that Boys should wear blue T-Shirts and Girls should wear pink T-Shirts. If a boy was given a pink T-Shirt or a girl was given a Blue T-Shirt, he called it an inversion. So, Xenny wanted to distribute T-Shirts in the above-mentioned interesting manner and also wanted to minimize "inversions". Help him solve the task. Note: There are no disjoint groups of friends in the room. That is, 2 distinct groups with finite number of students do not exist, but exactly 1 group of students exists in the given situation. Input The first line is the number of test cases T. First line of each test case contains 2 space-separated integers - N and M - number of students and number of friendships present respectively. Second line consists of N space-separated characters, where ith character denotes the gender of the ith student. B: Boy, G: Girl. M lines follow. Each line consists of 2 space-separated integers, u and v, showing that u is a friend of v and vice-versa. Output If Xenny could distribute the T-Shirts in the desired way, print the minimum number of inversions required. Else, print -1. Constraints 1 ≤ N ≤ 105 1 ≤ M ≤ 105 1 ≤ u, v ≤ N Colors of T-Shirt are represented by uppercase characters 'B' and 'G' Sample Input 3 3 2 B G B 1 2 1 3 6 9 B B B G G G 3 5 2 6 4 2 6 3 3 1 3 4 6 1 5 1 1 4 6 5 G G G B G G 6 3 1 3 2 3 4 3 5 3 Output 1 -1 2 Explanation #1 Student 1 can be given a Blue T-Shirt. Hence, Student 2 and 3 would receive Pink T-Shirts. Since, Student 3 is a Boy and has received a Pink T-Shirt, number of inversions = 1. #include<iostream> using namespace std; #define pii pair<int,int> #define piii pair<pii,int> const int mn =1e5+6; const int maxS = 2e6+6; template <typename T> class Vector{ private: int size; int capacity; T *array; void expand(int newCapacity) { if (newCapacity <= size) return; T * old = array; array = new T[newCapacity]; for (int i = 0; i < size; i++) array[i] = old[i]; delete[] old; capacity = newCapacity; } public: Vector(int initCapacity = 1) { size = 0; capacity = initCapacity; array = new T[capacity]; } ~Vector() { delete[] array; } Vector & operator=(Vector & rhs) { if (this != &rhs) { delete[] array; size = rhs.size; capacity = rhs.capacity; array = new T[capacity]; for (int i = 0; i < size; i++) array[i] = rhs.array[i]; } return *this; } int Size() { return size; } T & operator[](int index) { return array[index]; } void push_back(T newElement) { if (size == capacity) expand(2 * size); array[size] = newElement; size++; } void popBack() { size--; } void clear() { size = 0; } }; Vector<int> g[mn]; char a[mn]; int no[mn],dx[mn],dy[mn],xet[mn]; int n,m,dem1,dem2,res,check; void dfs(int x) { for(int i=0;i<g[x].Size();i++) { int y=g[x][i]; if(xet[y]==xet[x]){check=1;break;} if(xet[y]!=0)continue; if(check==0) { xet[y]=3-xet[x]; if(xet[y]==1&&a[y]=='G')dem1++; if(xet[y]==2&&a[y]=='B')dem1++; if(xet[y]==1&&a[y]=='B')dem2++; if(xet[y]==2&&a[y]=='G')dem2++; dfs(y); } } } int main() { ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0); //freopen("1.txt","r",stdin); //freopen(".out","w",stdout); int te=10; cin>>te; for(int tes=1;tes<=te;tes++) { //cout<<"Case "<<tes<<"\n"; res=0; check=0; cin>>n>>m; for(int i=1;i<=n;i++){ cin>>a[i]; g[i].clear(); xet[i]=0; } for(int i=1;i<=m;i++) { int x,y; cin>>x>>y; g[x].push_back(y); g[y].push_back(x); } for(int i=1;i<=n;i++) if(xet[i]==0) { dem1=dem2=0; xet[i]=1; if(xet[i]==1&&a[i]=='G')dem1++; if(xet[i]==2&&a[i]=='B')dem1++; if(xet[i]==1&&a[i]=='B')dem2++; if(xet[i]==2&&a[i]=='G')dem2++; dfs(i); if(dem1<dem2)res+=dem1; else res+=dem2; } //for(int i=1;i<=n;i++)cout<<xet[i]<<" "; if(check)cout<<-1<<'\n'; else cout<<res<<'\n'; } return 0; }

Leave a Comment