Bitmasking

Author: CodeRex7

geekforgeeks/BitMasking_&_DP/Unique_cap.cpp

@@ -0,0 +1,63 @@
+/*
+There are 100 different types of caps each having a unique id from 1 to 100.
+Also, there are ‘n’ persons each having a collection of a variable number of caps.
+One day all of these persons decide to go in a party wearing a cap but to look unique
+they decided that none of them will wear the same type of cap.
+So, count the total number of arrangements or ways such that none of them is wearing the same type of cap.
+Since, number of ways could be large, so output modulo 1000000007 .
+*/
+#include<bits/stdc++.h>
+#define MOD 1000000007
+using namespace std;
+
+vector<int>capList[101];
+int dp[1025][101];
+int allmask;
+
+long long int CountWaysUtil(int mask, int i){
+	if(mask == allmask)
+		return 1;
+	if(i>100)
+		return 0;
+	if(dp[mask][i] != -1)
+		return dp[mask][i];
+
+	long long int ways = CountWaysUtil(mask,i+1);
+	int size = capList[i].size();
+	for(int j=0;j<size;j++){
+		if(mask & (1<<capList[i][j]))
+			continue;
+		else 
+			ways+= CountWaysUtil(mask | (1<<capList[i][j]), i+1);
+		ways%=MOD;
+	}
+	return dp[mask][i]=ways;
+}
+
+void CountWays(int n){
+	string temp,str;
+	int x;
+	for(int i=0;i<n;i++){
+		getline(cin,str);
+		stringstream ss(str);
+		while(ss >> temp)
+		{
+			stringstream s;
+			s << temp;
+			s >> x;
+			capList[x].push_back(i);
+ 		}
+	}
+	allmask = (1<<n)-1;
+	memset(dp,-1,sizeof dp);
+	cout<<CountWaysUtil(0,1)<<endl;
+}
+
+int main()
+{
+	int n;
+	cin>>n;
+	cin.ignore();
+	CountWays(n);
+	return 0;
+}