// Bitstrings as a Service // Solution by Jacob Plachta #define DEBUG 0 #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define LL long long #define LD long double #define PR pair #define Fox(i,n) for (i=0; 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 T Abs(T x) { return(x<0 ? -x : x); } template 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 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); }