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16 Partition Problem.cpp
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16 Partition Problem.cpp
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/**
Problem : Partition Problem
Given a set of positive integers, find if it can be divided into two subsets with equal sum.
**/
/**Which of the favors of your Lord will you deny ?**/
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define PII pair<int,int>
#define PLL pair<LL,LL>
#define MP make_pair
#define F first
#define S second
#define INF INT_MAX
#define ALL(x) (x).begin(), (x).end()
#define DBG(x) cerr << __LINE__ << " says: " << #x << " = " << (x) << endl
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class TIn>
using indexed_set = tree<
TIn, null_type, less<TIn>,
rb_tree_tag, tree_order_statistics_node_update>;
/*
PBDS
-------------------------------------------------
1) insert(value)
2) erase(value)
3) order_of_key(value) // 0 based indexing
4) *find_by_order(position) // 0 based indexing
*/
inline void optimizeIO()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
const int nmax = 2e3+7;
const LL LINF = 1e17;
string to_str(LL x)
{
stringstream ss;
ss<<x;
return ss.str();
}
//bool cmp(const PII &A,const PII &B)
//{
//
//}
vector<int> ara = { -1,7,3,1,5,4,8};
//vector<int> ara = { -1,3,1,1,2,2,1};
/** 1 based indexing **/
/**
dp[i][j] = include or exclude
base case : if sum possible return true
**/
int dp[nmax][nmax];
int solve(int i,int s)
{
if(i<=0 || s<0)
return 0;
if(s==0)
return 1;
int &ret = dp[i][s];
if(ret!=-1)
return ret;
int inc = solve(i-1,s-ara[i]);
int exc = solve(i-1,s);
return ret = inc|exc;
}
string solve_print(int i,int s)
{
if(i==0 || s<0)
return "";
if(s==0)
return "";
int inc = solve(i-1,s-ara[i]);
int exc = solve(i-1,s);
if(inc)
return solve_print(i-1,s-ara[i]) + " " + to_str(i);
else
return solve_print(i-1,s);
}
int main()
{
optimizeIO();
memset(dp,-1,sizeof dp);
int n = ara.size()-1;
int total = 0;
for(int i=1;i<=n;i++)
total += ara[i];
if(total%2==1)
cout<<"Not Possible"<<endl;
else
{
cout<<solve(n,total/2)<<endl;
cout<<"One part indexes : "<<solve_print(n,total/2)<<endl;
}
// for(int i=1; i<=n; i++)
// {
// for(int j=1; j<=max_cap; j++)
// cout<<dp[i][j]<<" ";
// cout<<endl;
// }
return 0;
}