Exemplo n.º 1
0
  int count(vector<string> S, int K) {
    int N=SZ(S),M=0; REP(i,N)M+=SZ(S[i]);
    VI P(N); REP(i,N)P[i]=i;
    int res=0;

    VI divs;
    if(M!=K) divs.push_back(1);
    int L=M/K;
    for(int p=2; p*p<=L; ++p) if(L%p==0) {
      divs.push_back(p);
      if(p!=L/p) divs.push_back(L/p);
    }
//    dumpAR(divs);
    
    do {
      string s;
      REP(i,N) s+=S[P[i]];
      if(M%K!=0) return 0;
      
//      dump2(s,cycle(s,M/K));
      
      if(!cycle(s,M/K)) continue;
      
      bool cy=false;
      FORR(d,divs) {
        cy|=cycle(s.substr(0,M/K),d);
      }
      if(cy) continue;
//      dump(s);
      ++res;
    } while(next_permutation(P.begin(),P.end()));
Exemplo n.º 2
0
int main(int argc, char** argv)
{
    constexpr int default_n = 100;

    int n = argc > 1 && atoi(argv[1]) > default_n ? atoi(argv[1]) : default_n;

    VI va;
    VI vb;

    for (int i = 0; i < n; ++i)
        va.push_back(generate_x(i + 1));

    for (int i = 0; i < n; ++i)
        vb.push_back(generate_x(i + 1));

    sort(va.begin(), va.end());
    sort(vb.begin(), vb.end());

    auto start_time = steady_clock::now();

    auto r1 = find_common(va, vb);

    cout << "use " << duration_cast<microseconds>(steady_clock::now() - start_time).count() << " microseconds\n";
    start_time = steady_clock::now();

    auto r2 = find_common1(va, vb);
    cout << "use " << duration_cast<microseconds>(steady_clock::now() - start_time).count() << " microseconds\n";

    if (r1 != r2) cout << "error\n";

    return 0;
}
Exemplo n.º 3
0
void find_common_helper2(Iter s1, Iter e1, Iter s2, Iter e2, VI& result)
{
    if (s1 == e1 || s2 == e2) return;

    if (*s1 == *s2)  {
        result.push_back(*s1);
        ++s1, ++s2;
    } else {
        if (*s1 > *s2) {
            swap(s1, s2);
            swap(e1, e2);
        }

        s1 = lower_bound(s1, e1, *s2);
    }

    if (s1 == e1 || s2 == e2) return;

    auto l1 = e1 - 1;
    auto l2 = e2 - 1;
    if (*l1 == *l2)  {
        result.push_back(*l1);
        --e1, --e2;
    } else {
        if (*l1 < *l2) {
            swap(s1, s2);
            swap(e1, e2);
        }

        l1 = lower_bound(s1, e1, *e2);
        e1 = l1 + 1;
    }

    find_common_helper(s1, e1, s2, e2, result);
}
bool isOk(VI a, VI b) {
    memset(ok, 0, sizeof(ok));
    if(binary_search(a.begin(), a.end(), 6)) {
        a.push_back(9);
    }
    if(a.back()==9) {
        a.push_back(6);
    }
    if(binary_search(b.begin(), b.end(), 6)) {
        b.push_back(9);
    }
    if(b.back()==9) {
        b.push_back(6);
    }
    VI::iterator i, j;
    for(i=a.begin(); i!=a.end(); ++i) {
        for(j=b.begin(); j!=b.end(); ++j) {
            ok[(*i)*10+(*j)]=true;
            ok[(*j)*10+(*i)]=true;
        }
    }
    bool ret=true;
    for(int in=1; in<10; ++in) {
        if(!ok[in*in]) {
            ret=false;
        }
    }
    return ret;
}
Exemplo n.º 5
0
void stableMatching (int n, VVI& maleRank, VVI& femaleRank, VI& wife) {
    // a male m prefers w to w' if maleRank[m][w] < maleRank[m][w']
    // returns male-optimal matching

    VI freeMen;
    VVPI fq(n);
    VI husband(n, -1);
    for (int m = 0; m < n; ++m) {
        for (int w = 0; w < n; ++w) {
            fq[m].push_back(make_pair(maleRank[m][w], w));
        }
        sort(all(fq[m]), greater<PI>());
        freeMen.push_back(m);
    }
    while (!freeMen.empty()) {
        int m = freeMen.back(), w = fq[m].back().y;
        fq[m].pop_back();
        if (husband[w] == -1) {
            husband[w] = m;
            freeMen.pop_back();
        } else if (femaleRank[w][m] < femaleRank[w][husband[w]]) {
            freeMen.pop_back();
            freeMen.push_back(husband[w]);
            husband[w] = m;
        }
    }
    wife = VI(n);
    for (int w = 0; w < n; ++w) {
        wife[husband[w]] = w;
    }
}
Exemplo n.º 6
0
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        // Init
        VI ans;
        VVI adj(n);
        QI q;
        VI d(n, -1), pathTo(n);
        for(auto& pr: edges) {
            int u = pr.first, v = pr.second;
            adj[u].push_back(v);
            adj[v].push_back(u);
        }
        // Find one endpoint of longest path
        int start = 0;
        q.push(0);
        d[0] = 0;
        while(!q.empty()) {
            int u = q.front(); q.pop();
            start = u;
            
            for(int v: adj[u]) {
                if(d[v] == -1) {
                    d[v] = d[u] + 1;
                    q.push(v);
                }
            }
        }

        // Find longest path
        int maxd = 0, end = start;
        q = QI();
        fill(d.begin(), d.end(), -1);
        q.push(start);
        d[start] = 0;
        while(!q.empty()) {
            int u = q.front(); q.pop();
            maxd = d[u];
            end = u;

            for(int v: adj[u]) {
                if(d[v] == -1) {
                    pathTo[v] = u;
                    d[v] = d[u] + 1;
                    q.push(v);
                }
            }
        }

        // Find mid points
        int s = end;
        for(int i = 0; i < maxd / 2; i++) {
            s = pathTo[s];
        }
        ans.push_back(s);
        if(maxd & 1) {
            ans.push_back(pathTo[s]);
        }

        return ans;
    }
Exemplo n.º 7
0
	VI getDigits(int n) {
		VI result;
		while (n > 9) {
			result.push_back(n % 10);
			n /= 10;
		}
		result.push_back(n);
		return result;
	}
Exemplo n.º 8
0
void findSize(FILE *file1, FILE *file2, int &row, int &col, VI &len1, VI &len2)
{
	while(!feof(file1) || !feof(file2)) {
		fgets(in1, SMAX, file1);
		fgets(in2, SMAX, file2);
		if (!feof(file1)) { row++; len1.push_back(strlen(in1));}
		if (!feof(file2)) { col++; len2.push_back(strlen(in2));}
	}
}
Exemplo n.º 9
0
bool solve(int n, int k) {
   if (k == 1)
      return true;
   if (k > MAXK)
      return false;
   if (fact[k] > n)
      return false;

   VI v;
   int N = n;
   for (VI::iterator it = primes.begin(); it != primes.end() && (*it) * (*it) <= N; ++it)
      if (N % *it == 0) {
         v.push_back(0);
         while (N % *it == 0) {
            ++v.back();
            N /= *it;
         }
      }
   if (N > 1)
      v.push_back(1);
   sort(v.begin(), v.end());
   
   reverse(v.begin(), v.end());
   k -= (int) v.size() + 1;
   for (VI::iterator it = v.begin(); it != v.end(); ++it)
      *it -= 1;
   while (!v.empty() && v.back() == 0) {
      v.pop_back();
   }

   if (k <= 0)
      return true;

   if (DBG + 0) {
      cerr << "BEGIN!!!!!\n";
      for (VI::iterator it = v.begin(); it != v.end(); ++it)
         cerr << *it << " -- ";
      cerr << endl;
      cerr << k << endl;
   }
   VI w(v.size(), 1);
   bool found = walk1(v, w, 0, k/*,set<VI>()*/);

   if (!found) {
      if (v.size() >= 3 && v[2] >= 3) {
         v[0] -= 1;
         v[1] -= 1;
         v[2] -= 1;
         found = walk1(v, w, 0,  k - 1/*,set<VI>()*/);
      }
   }

   return found;
}
Exemplo n.º 10
0
 vector<int> findPoint(int x1, int y1, int x2, int y2) {
   VI res;
   for(int i=-100; i<=100; ++i) for(int j=-100; j<=100; ++j) {
     if(i!=x1&&i!=x2&&j!=y1&&j!=y2&&dist(x1,y1,i,j)>dist(x2,y2,i,j)) {
       res.push_back(i),res.push_back(j);
       return res;
     }
   }
   
   return res;
 }
Exemplo n.º 11
0
 //lastrow = row for the last column
 void getPath(const VVI &parent, int lastrow, VI &path)
 {
    int nc = parent.size();
    path.clear();
    path.push_back(lastrow);
    if(nc > 1){
       for(int c = nc - 2; c >= 0; c--){
          //decide row for column c
          int r = parent[c+1][path.back()];
          path.push_back(r);
       }
    }
 }
Exemplo n.º 12
0
int main()
{
    scanf("%d",&n);
    for (int i=1;i<n;++i){
        int a,b;
        scanf("%d%d",&a,&b);
        --a;--b;
        adj[a].PB(b);
        adj[b].PB(a);
    }
    
    int root=rand()%n;
    dfs(root,-1);
    
    memset(ans,false,sizeof(ans));
    for (int i=0;i<n;++i){
        VI a;
        if (i!=root) a.push_back(size[root]-size[i]);
        for (int j=0;j<adj[i].size();++j){
            int v=adj[i][j];
            if (v!=father[i]){
                a.push_back(size[v]);
            }
        }
        int s=0;
        memset(f,false,sizeof(f));
        f[0]=true;

        for (int j=0;j<a.size();++j){
            s+=a[j];
            for (int k=s;k>=a[j];--k)
                f[k]|=f[k-a[j]];
        }
        for (int j=1;j<s;++j)
            ans[j]|=f[j];
    }
    
    int cnt=0;
    for (int i=1;i<n-1;++i) cnt+=ans[i];
    printf("%d\n",cnt);
    for (int i=1;i<n-1;++i)
    if (ans[i]) printf("%d %d\n",i,n-1-i);
    
    if (cnt==0){
        VI a;
        printf("%d\n",a[0]);
    }
    
    return 0;
}
Exemplo n.º 13
0
 int minimumQueries(vector<string> GG) {
   int R=SZ(GG),C=SZ(GG[0]);
   ZERO(viz);
   REP(i,100) G[i].clear();
   //    REP(r,R) G[r].push_back(R+r),G[R+r].push_back(r);
   REP(i,R) REP(j,C) if(GG[i][j]=='Y') {
     int u=i,v=R+j;
     G[u].push_back(v),G[v].push_back(u);
   }
   int res=0;
   VI temp;
   REP(u,R) if(!viz[u]) ++res,dfs(u),temp.push_back(u);
   REP(u,C) if(!viz[u+R]) ++res,dfs(u+R),temp.push_back(u);
   //     dumpAR(temp);
   return res-1;
 }
Exemplo n.º 14
0
    int DFS (PI v, int index) {
        idx[v.x] = index;
        low[v.x] = index;
        index += 1;

        int children = 0;
        bool ap = false;
        for (auto w : adj[v.x]) if (w.y != v.y) {
                if (idx[w.x] == -1) {
                    index = DFS(w, index);
                    low[v.x] = min(low[v.x], low[w.x]);
                    if (low[w.x] > idx[v.x]) {
                        bridges.push_back(mp(min(w.x, v.x), max(w.x, v.x)));
                    }
                    children++;
                    if ((v.y == -1 && children >= 2) || (v.y != -1 && low[w.x] >= idx[v.x])) {
                        ap = true;
                    }
                } else {
                    low[v.x] = min(low[v.x], idx[w.x]);
                }
            }
        if (ap) {
            cutVertices.push_back(v.x);
        }
        return index;
    }
Exemplo n.º 15
0
	vector <int> reconstruct(string signature) {
		VI res; res.clear();
        res.push_back(1);
        int mxptr = 0;
        REP(i, SZ(signature)){
            if(signature[i] == 'I') {
                res.push_back(i+2);
                mxptr = SZ(res) - 1;
            }
            else{
                FOR(j, mxptr, SZ(res)) res[j]++;
                res.push_back(res[SZ(res)-1] - 1);
            }
        }
        return res;
	}
Exemplo n.º 16
0
void run(int u, int l, VI p, vector<VI> used){
	vector<VI> v;
	VI np;
	if(l==8){
		for(int i=0; i<p.size(); i++){
			printf("%d",p[i]+1);
		}
		printf("\n");
	}
	else{
		for(int j=0; j<5; j++){
			if(m[u][j]){
				if(!used[u][j]){
					v.assign(5,VI());
					np = p;
					np.push_back(j);
					for(int i=0; i<5; i++)
						for(int j=0; j<5; j++)
							v[i].push_back(used[i][j]);
					v[u][j] = 1;
					v[j][u] = 1;
					run(j,l+1,np,v);
				}
			}
		}
	}
}
Exemplo n.º 17
0
VI LongestIncreasingSubsequence(VI v)
{
    VPII best;
    VI dad(v.size(), -1);

    for (int i = 0; i < v.size(); i++)
    {
#ifdef STRICTLY_INCREASNG
        PII item = make_pair(v[i], 0);
        VPII::iterator it = lower_bound(best.begin(), best.end(), item);
        item.second = i;
#else
        PII item = make_pair(v[i], i);
        VPII::iterator it = upper_bound(best.begin(), best.end(), item);
#endif
        if (it == best.end())
        {
            dad[i] = (best.size() == 0 ? -1 : best.back().second);
            best.push_back(item);
        }
        else
        {
            dad[i] = dad[it->second];
            *it = item;
        }
    }

    VI ret;
    for (int i = best.back().second; i >= 0; i = dad[i])
        ret.push_back(v[i]);
    reverse(ret.begin(), ret.end());
    return ret;
}
Exemplo n.º 18
0
int main(){
	int nk, np;
	while(cin >> nk >> np && nk!=0 && np!=0){
		MaxFlowDinic D(nk+np+5);
		VI aux;
		int R = 0;
		for(int i=0; i<nk; i++){
			int c;
			cin >> c;
			R += c;
			aux.push_back(c);
			D.add_edge(0, i+1, c);
		}
		for(int i=0; i<np; i++){
			int k;
			cin >> k;
			for(int j=0; j<k; j++){
				int ct;
				cin >> ct;
				D.add_edge(ct, i+nk+1, INF);
			}
			D.add_edge(i+nk+1, nk+np+1);
		}
		
		int d = D.dinic(0, nk+np+1);
		bool an = true;
		
		int res[np];
		memset(res, 0, sizeof(res));
		for(int i=0; i<D.edges.size(); i+=2){
			if(D.edges[i].u == 0 || D.edges[i].v == nk+np+1) continue;
			if(D.edges[i].flow <= 0) continue;
			if(D.edges[i].v <= nk) continue;
			res[ D.edges[i].v-1-nk ] = D.edges[i].u;
		}
		for(int i=1; i<=nk; i++){
			int r = 0;
			for(int j=0; j<np; j++){
				if(res[j] == i){
					r++;
				}
			}
			if(r != aux[i-1]) an = false;
		}
		
		if(an && d == R){
			cout << 1 << endl;
			for(int i=1; i<=nk; i++){
				for(int j=0; j<np; j++){
					if(res[j] == i){
						cout << j+1 << " ";
					}
				}
				cout << endl;
			}
		}else{
			cout << 0 << endl;
		}
	}
}
Exemplo n.º 19
0
    void DFS (int v) {
        idx[v] = index;
        low[v] = index;
        index += 1;
        st.push_back(v);
        inStack[v] = true;

        for (auto w : adj[v]) {
            if (idx[w] == -1) {
                DFS(w);
                low[v] = min(low[v], low[w]);
            } else if (inStack[w]) {
                low[v] = min(low[v], low[w]);
            }
        }

        if (low[v] == idx[v]) {
            int w;
            components.push_back(VI());
            do {
                w = st.back();
                st.pop_back();
                inStack[w] = false;
                componentOf[w] = totalComponents;
                components[totalComponents].push_back(w);
            } while (w != v);
            totalComponents++;
        }
    }
Exemplo n.º 20
0
int main()
{
    int first=1,m,i,j,len,dlen;
    char cmd[15],oj,tc;
    while(1)
    {
        if(first) first=0;
        else scanf("%c%c",&tc,&tc);
        if(gets(a)==NULL) break;
        pos.clear();ans=0;
        gets(a);
        gets(b);
        len = strlen(b);
        alen = strlen(a);
        dlen = alen - len + 1;
        //findallsub();
        for(i=0;i<dlen;i++)
        {
            for(j=0;;j++)
            {
                if(b[j]=='\0')
                {
                    pos.push_back(i);
                    ans++;
                    break;
                }
                if(b[j]!=a[i+j]) break;
            }
        }
//        show();
        //-------------
        scanf("%d",&m);
        while(m--)
        {
            //printf("i=%d\n",m);
            scanf("%s",cmd);
            //printf("%s\n",cmd);
            if(ans==0)
            {
                if(strcmp(cmd,"Query")==0)
                    printf("0\n");
                continue;
            }
            if(strcmp(cmd,"Query")==0)
                printf("%d\n",ans);
            else
            {
                scanf("%c%c",&tc,&oj);
                if(strcmp(cmd,"Push_back")==0)
                    update(oj,len);
                else
                    update(oj,-1);
                len++;
//                show();
            }
        }
    }
    return 0;
}
Exemplo n.º 21
0
bool can(int a, VI v){
    VI tmp;
    REP(i, SZ(v)) if(a%v[i] == 0) tmp.push_back(v[i]);
    if(SZ(tmp) == 0) return false;
    int l = tmp[0];
    FOR(i, 1, SZ(tmp)) l = lcm(l, tmp[i]);
    return l == a;
}
Exemplo n.º 22
0
void valid(int d,int c,int i,long n){
  if(c==0) {cout<<n<<endl;v.push_back(n);}
  else {
    for(int j=i;j<=10-c;j++) {
      valid(d,c-1,j,(n+(long)(j*pow(10,d-c))));
    }
  }
}
Exemplo n.º 23
0
int main(int argc, char *argv[]) {
	int TC, px, py;
	double ans;
	scanf("%d",&TC);
	while(TC-- > 0) {
		scanf("%d%d",&S,&P);
		X.clear();
		Y.clear();
		for(int i = 0; i < P; ++i) {
			scanf("%d%d",&px,&py);
			X.push_back(px);
			Y.push_back(py);
		}
		ans=binary_search();
		printf("%.2lf\n",ans);
	}
	return 0;
}
Exemplo n.º 24
0
 string isPossible(vector<int> V) {
   VI xs;
   REP(i,SZ(V)-1) xs.push_back(V[i+1]/V[i]);
   sort(xs.begin(),xs.end());
   
   REP(i,SZ(xs)) {
     int cnt=i+1;
     if(xs[i]<=cnt) return no;
   }
Exemplo n.º 25
0
void solve()
{
    int n, m; cin >> n >> m;
    VI a(m); for(int i=0; i<m; ++i) cin >> a[i];
    VI b(m);
    int sum = n;
    while(sum) {
        int minval=0, cnt=0;
        for(int i=0; i<m; ++i) if (a[i]>0)
        { cnt++; if (minval==0||a[i]<minval) minval=a[i]; }

        if (minval==0) break;

        if (sum/cnt==0) {
            for(int i=0; i<m; ++i) if (a[i]>0)
            { a[i]--; b[i]++; sum--; if (sum==0) break;}
        }
        else {
            int decr = min(sum/cnt, minval);
            for(int i=0; i<m; ++i) if (a[i]>0)
            { a[i]-=decr; b[i]+=decr; sum-=decr; }
        }
    }
    if (sum >0) { cout << -1 << endl; return; }
    int maxval=0,total=0;
    for(int i=0; i<m; ++i)
    { maxval = max(maxval, b[i]); total += b[i]; }
    if (maxval > total-maxval) { cout << -1 << endl; return; }

    vector<PII> avail;
    for(int i=0; i<m; ++i) if (b[i]) avail.push_back(PII(b[i],i));
    sort(avail.begin(), avail.end()); reverse(avail.begin(), avail.end());
    VI ans;
    while(avail[0].first) {
        for(int x=0; x<int(avail.size()) && avail[x].first; ++x)
        {
            int pp = (avail[x].second+1); ans.push_back(pp);
            avail[x].first--;
        }
    }
    int kk = ans.size(); 
    while(ans[kk-1]==ans[0])
    {
        for(int i=1; i<kk; ++i) if (ans[i]!=ans[0]&&ans[i-1]!=ans[0])
        {
            VI tmp=ans;
            for(int j=i; j<kk-1; ++j) tmp[j+1] = ans[j];
            tmp[i] = ans[kk-1];
            ans = tmp;
            break;
        }
    }
    cout << ans[0];
    for(int i=1; i<kk; ++i) cout << " " << ans[i];
    cout << endl;
}
	VI find(int T, VI requiredTime) {
		sort(requiredTime.begin(), requiredTime.end());
		int past = 0;
		int solved = 0;
		int penalty = 0;
		VI::const_iterator it;
		for (it = requiredTime.begin(); it != requiredTime.end(); ++it) {
			past += *it;
			if (past > T) {
				break;
			}
			++solved;
			penalty += past;
		}
		VI r;
		r.push_back(solved);
		r.push_back(penalty);
		return r;
	}
Exemplo n.º 27
0
// finds all solutions to ax = b (mod n)
VI modular_linear_equation_solver(int a, int b, int n) {
  int x, y;
  VI solutions;
  int d = extended_euclid(a, n, x, y);
  if (!(b%d)) {
    x = mod (x*(b/d), n);
    for (int i = 0; i < d; i++)
      solutions.push_back(mod(x + i*(n/d), n));
  }
  return solutions;
}
Exemplo n.º 28
0
void sieve() {
   VI P(MAXP, 1);
   P[0] = P[1] = 0;
   for (int i = 2; i < MAXP; ++i)
      if (P[i]) {
         primes.push_back(i);
         for (int x = i * i; x < MAXP; x += i)
            P[x] = 0;
      }

}
Exemplo n.º 29
0
int main(int argc, char const *argv[]) {
    ios::sync_with_stdio(false);

    TarjanSCC *T;
    TransitiveRelation *Tr;

    int t;
    cin >> t;
    for (int cs = 0; cs < t; ++cs) {
        int n, m;
        cin >> n >> m;
        T = new TarjanSCC(n);
        for (int i = 0; i < m; ++i) {
            int a, b;
            cin >> a >> b;
            T->addEdge(a - 1, b - 1);
        }
        T->buildSCC();
        Tr = new TransitiveRelation(T->totalComponents);
        int maxr = 0, minr = 0;

        for (int i = 0; i < T->totalComponents; ++i) {
            VB flag(T->totalComponents, false);

            for (auto u : T->components[i]) {
                for (auto v : T->adj[u]) {
                    flag[T->componentOf[v]] = true;
                }
            }

            for (int j = i - 1; j >= 0; --j) if (flag[j]) {
                    Tr->addEdge(i, j);
                }

            maxr += T->components[i].size() * (T->components[i].size() - 1);
            if (T->components[i].size() >= 2) {
                minr += T->components[i].size();
            }
        }

        VI w;
        for (auto c : T->components) {
            w.push_back(c.size());
        }

        PI res = Tr->getReductionAndClosure(w);
        minr += res.x, maxr += res.y;

        cout << "Case #" << cs + 1 << ": " << minr << ' ' << maxr << endl;
    }

    return 0;
}
Exemplo n.º 30
0
bool chk(string s, VI c, int b, int w){
	int n = SZ(s);
	int ptr = 0;
	VI remain;
	int now_w = 0;
	
	while(ptr < n){
		if(s[ptr] >= '0' && s[ptr] <= '2') c[s[ptr]-'0']--, ptr++;
		else{
			int i = ptr;
			int cnt = 0;
			while(i < n && s[i] == 'b') cnt++, i++;
			if(cnt & 1){
				now_w++;
				if(now_w == w) return false;
				if(ptr == 0){
					if(cnt == 1) return false;
					else remain.push_back(1);
				}
				else remain.push_back(ptr-1);
				b -= (cnt+1)/2;
			}
			else b -= cnt/2;
			ptr = i; 
		}
	}
	REP(i, 3) if(c[i] < 0) return false;
	if(b < 0) return false;
	
	int t = 0;
	REP(i, 3) t += c[i];
	
	REP(i, SZ(remain)) while(remain[i] > 0){
		if(remain[i] >= 2 && b > 0) remain[i] -= 2, b--;
		else t--, remain[i]--;
		if(t < 0) return false;
	}
	
	return true;
}