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| // Bitstrings as a Service | |
| // Solution by Jacob Plachta | |
| using namespace std; | |
| 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); | |
| 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); | |
| } | |
| 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); | |
| } |