Black and White Tree

Problem Statement :

```Nikita is making a graph as a birthday gift for her boyfriend, a fellow programmer! She drew an undirected connected graph with N nodes numbered from 1 to N in her notebook.

Each node is shaded in either white or black. We define nW to be the number of white nodes, and nB to be the number of black nodes. The graph is drawn in such a way that:

No 2 adjacent nodes have same coloring.
The value of |nW - nB|, which we'll call D, is minimal.
Nikita's mischievous little brother erased some of the edges and all of the coloring from her graph! As a result, the graph is now decomposed into one or more components. Because you're her best friend, you've decided to help her reconstruct the graph by adding K edges such that the aforementioned graph properties hold true.

Given the decomposed graph, construct and shade a valid connected graph such that the difference |nW - nB| between its shaded nodes is minimal.

Input Format

The first line contains 2 space-separated integers, N (the number of nodes in the original graph) and M (the number of edges in the decomposed graph), respectively.
The M subsequent lines each contain2  space-separated integers, u and v, describing a bidirectional edge between nodes u and v in the decomposed graph.

Constraints
1 <= N <= 2*10^5
0 <= M <= min(5*10^5, (N *(N-1))/2)

It is guaranteed that every edge will be between 2 distinct nodes, and there will never be more than 1 edge between any 2 nodes.
The graph is connected and no 2 adjacent nodes have the same coloring.
The value of |nB - nW| is minimal.
K <= 2*10^5

Output Format

You must have K+1 lines of output. The first line contains 2 space-separated integers: D (the minimum possible value of |nB - nW|) and K (the number of edges you've added to the graph), respectively.
Each of the K subsequent lines contains 2 space-separated integers, u and v, describing a newly-added bidirectional edge in your final graph (i.e.: new edge u <-> v).

You may print any 1 of the possible reconstructions of Nikita's graph such that the value of D in the reconstructed shaded graph is minimal.```

Solution :

```                            ```Solution in C :

In C++ :

#include <bits/stdc++.h>
using namespace std;

#define ft first
#define sd second
#define pb push_back
#define all(x) x.begin(),x.end()

#define ll long long int
#define vi vector<int>
#define vii vector<pair<int,int> >
#define pii pair<int,int>
#define vl vector<ll>
#define vll vector<pair<ll,ll> >
#define pll pair<ll,ll>
#define pli pair<ll,int>
#define mp make_pair

#define sc1(x) scanf("%d",&x)
#define sc2(x,y) scanf("%d%d",&x,&y)
#define sc3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define scll1(x) scanf("%lld",&x)
#define scll2(x,y) scanf("%lld%lld",&x,&y)
#define scll3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)

#define pr1(x) printf("%d\n",x)
#define pr2(x,y) printf("%d %d\n",x,y)
#define pr3(x,y,z) printf("%d %d %d\n",x,y,z)

#define prll1(x) printf("%lld\n",x)
#define prll2(x,y) printf("%lld %lld\n",x,y)
#define prll3(x,y,z) printf("%lld %lld %lld\n",x,y,z)

#define pr_vec(v) for(int i=0;i<v.size();i++) cout << v[i] << " " ;

#define f_in(st) freopen(st,"r",stdin)
#define f_out(st) freopen(st,"w",stdout)

#define fr(i, a, b) for(i=a; i<=b; i++)
#define fb(i, a, b) for(i=a; i>=b; i--)

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const int N=200010;
int w[2];
vector <int> v[N];
int a[N], d[N], p[N], c[N], col[N], sz, ccol[2][N];
bool f[N];
queue<int> q[N];
void dfs(int x,int y)
{
f[x]=1;
w[y]++;
col[x] = y;
ccol[y][sz] = x;
for (int i=0;i<v[x].size();i++)
if (!f[v[x][i]])
dfs(v[x][i],(y^1));
else if( col[v[x][i]] != 1 - y )
assert(0);
}

int main()
{
int n,m;
scanf("%d%d",&n,&m);
for (int i=0,x,y;i<m;i++)
{
scanf("%d%d",&x,&y);
v[x].push_back(y);
v[y].push_back(x);
}

int k=0;
for (int i=1;i<=n;i++)
{
if (!f[i])
{
w[0] = w[1]=0;
dfs(i,0);
c[i] = (w[0] < w[1]);
k+=abs(w[0]-w[1]);
a[abs(w[0]-w[1])]++;
q[abs(w[0]-w[1])].push( i );
}
}

for (int j=1;j<=k;j++)
d[j]=N;

for (int i=1;i<=n;i++)
{
if (a[i])
{
for (int j=0;j+i<=k;j++)
{
if( d[i+j] == N && d[j] != N )
{
d[i+j] = d[j] + 1;
p[i+j] = j;
}
}

for (int j=1;j<=k;j++)
{
if (d[j]>a[i])
{
d[j]=N;
p[j]=0;
}
else
{
d[j]=0;
}
}
}
}

int ans=k, v = 0;
for (int i=0;i<=k;i++)
{
if(d[i]<N)
{
if( ans >= abs(k - 2 * i) )
{
ans = abs(k - 2 * i);
v = i;
}
}
}

memset(f, 0, sizeof f);
while( v != 0 )
{
int diff = v - p[v];
int nd = q[diff].front();
q[diff].pop();
sz ++;
dfs(nd, c[nd]);
v = p[v];
}

for(int i=1; i<=n; i++)
{
if( !f[i] )
{
sz ++;
dfs(i, 1 - c[i]);
}
}

int blk, wht, idb, idw;
blk = wht = idb = idw = -1;
for(int i=1; i<=sz; i++)
{
if( ccol[0][i] )
{
blk = ccol[0][i];
idb = i;
}

if( ccol[1][i] )
{
wht = ccol[1][i];
idw = i;
}
}
cout << ans << " " << sz - 1 << "\n";
if( idb != idw ) cout << blk << " " << wht << "\n";
for(int i=1; i<=sz; i++)
{
if( i != idb && i != idw )
{
if( ccol[0][i] )
{
cout << wht << " " << ccol[0][i] << "\n";
}
else
{
cout << blk << " " << ccol[1][i] << "\n";
}
}
}
return 0;
}```
```

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