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hackercup / 2019 /round2 /bitstrings_as_a_service.cpp
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2019 Problems
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// Bitstrings as a Service
// Solution by Jacob Plachta
#define DEBUG 0
#include <algorithm>
#include <functional>
#include <numeric>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <complex>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <sstream>
using namespace std;
#define LL long long
#define LD long double
#define PR pair<int,int>
#define Fox(i,n) for (i=0; i<n; i++)
#define Fox1(i,n) for (i=1; i<=n; i++)
#define FoxI(i,a,b) for (i=a; i<=b; i++)
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
#define FoxR1(i,n) for (i=n; i>0; i--)
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
#define Min(a,b) a=min(a,b)
#define Max(a,b) a=max(a,b)
#define Sz(s) int((s).size())
#define All(s) (s).begin(),(s).end()
#define Fill(s,v) memset(s,v,sizeof(s))
#define pb push_back
#define mp make_pair
#define x first
#define y second
template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
template<typename T> T Sqr(T x) { return(x*x); }
string plural(string s) { return(Sz(s) && s[Sz(s)-1]=='x' ? s+"en" : s+"s"); }
const int INF = (int)1e9;
const LD EPS = 1e-12;
const LD PI = acos(-1.0);
#if DEBUG
#define GETCHAR getchar
#else
#define GETCHAR getchar_unlocked
#endif
bool Read(int &x)
{
char c,r=0,n=0;
x=0;
for(;;)
{
c=GETCHAR();
if ((c<0) && (!r))
return(0);
if ((c=='-') && (!r))
n=1;
else
if ((c>='0') && (c<='9'))
x=x*10+c-'0',r=1;
else
if (r)
break;
}
if (n)
x=-x;
return(1);
}
#define LIM 4001
int N,M,K;
int comp[LIM],sz[LIM],val[LIM];
vector<int> con[LIM];
int dyn[LIM][LIM];
void rec(int i)
{
if (comp[i]>=0)
return;
comp[i]=K;
sz[K]++;
int j;
Fox(j,Sz(con[i]))
rec(con[i][j]);
}
int main()
{
if (DEBUG)
freopen("in.txt","r",stdin);
// vars
int T,t;
int i,j;
// testcase loop
Read(T);
Fox1(t,T)
{
// init
Fox(i,N)
con[i].clear();
// input, and create graph of bit equalities
Read(N),Read(M);
while (M--)
{
Read(i),Read(j),i--,j--;
while (j>i)
{
con[i].pb(j);
con[j].pb(i);
i++,j--;
}
}
// find all components of equal bits
K=0;
Fill(sz,0);
Fill(comp,-1);
Fox(i,N)
if (comp[i]<0)
{
rec(i);
K++;
}
// DP on component value assignments
Fill(dyn,-1);
dyn[0][0]=0;
Fox(i,K)
Fox(j,N)
if (dyn[i][j]>=0)
dyn[i+1][j]=dyn[i+1][j+sz[i]]=j;
// trace back to assign values to components
Fox(i,N+1)
{
if (dyn[K][j=N/2+i]>=0)
break;
if (dyn[K][j=N/2-i]>=0)
break;
}
FoxR1(i,K)
{
val[i-1]=(dyn[i][j]==j ? 0 : 1);
j=dyn[i][j];
}
// output
printf("Case #%d: ",t);
Fox(i,N)
printf("%d",val[comp[i]]);
printf("\n");
}
return(0);
}