Linear brush title group recorded

HDU3949 XOR

• Linear substrate sub-topic
• Given the number of some
• These request numbers obtained by exclusive or a large value of k.
• Code:
  

   1 #include <bits/stdc++.h>
   2 #define nmax 10010
   3 
   4 using namespace std;
   5 typedef long long ll;
   6 ll a[nmax],b[70],k[70]; //k[i] 第i小的数
   7 int n,q;
   8 
   9 int main(){
  10     //freopen("owo.in","r",stdin);
  11     int T;
  12     cin>>T;
  13     for (int cas=1; cas<=T; cas++) {
  14         memset(b,0,sizeof(b));
  15         memset(k,0,sizeof(k));
  16         printf("Case #%d:\n",cas);
  17         scanf("%d",&n);
  18         for (int i=0; i<n; i++) {
  19             scanf("%lld",&a[i]);
  20             //求线性基
  21             for (int j=61 is ; J> = 0 ; J, ) {
   22 is                  IF (A [I] & (1LL << J)) {
   23 is                      IF (B [J]) A [I] = ^ B [J];
   24                      the else { B [J] = A [I]; BREAK ;}
   25                  }
   26 is              }
   27          }
   28          // up back-substitution, the way to count the number of 
  29          int CNT = 0 ;
   30          for ( int I = 0 ; I <= 61 is ; ++ I) IF (B [I]) {
   31 is                  K [CNT] = B [I];
   32                 cnt++;
  33                 for (int j=i+1; j<=61; j++) if(b[j]&(1LL<<i)) b[j]^=b[i];
  34         }
  35         ll mmax=(1LL<<cnt)-1;
  36         scanf("%d",&q);
  37         ll inq;
  38         bool flag=false;
  39         if(cnt<n) flag=true;
  40         for (int j=0; j<q; j++) {
  41             scanf("%lld",&inq);
  42             if(flag) inq--;
  43             if(inq>mmax) cout<<-1<<endl;
  44             else {
  45                 ll ans=0;
  46                 for (int i=0; (1LL<<i)<=inq; i++) if((1LL<<i)&inq) ans^=k[i];
  47                 printf("%lld\n",ans);
  48             }
  49         }
  50     }
  51     return 0;
  52 }

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BZOJ 2115: [Wc2011] Chorus

• The meaning of problems, an undirected graph, the node numbers 1 ~ n, find a path from 1 to n, edges and vertices may pass several times through the right side of seeking the maximum value of that path xor
• Solution: just find a path from 1 ~ n, and can be found in any path can be represented as a ring walk on this path, so find out all the rings
• Then the primary path is to find and exclusive OR largest in all of the rings in the
• Find all the rings dfs
• Code:
  

  . 1 #include <bits / STDC ++ H.>
   2  #define Nmax of 50010
   . 3  #define Mmax 100010
   . 4   
  . 5  the using  namespace STD;
   . 6 typedef Long  Long LL;
   . 7  int head [Nmax of] = { 0 }, D [Nmax of] = { 0 };   // is to sequentially vis 
  . 8 LL a [Nmax of Mmax +], tmp [Nmax of], B [ 70 ] = { 0 };
   . 9  int n-, m, IDX = 0 , IDXa = 0 ;   // idza is recording the number of ring 
  10  struct Edge {
   . 11     int v,ne;
  12     ll w;
  13 }e[mmax*2];
  14  
  15 inline void addedge(int u,int v,ll w){
  16     idx++;
  17     e[idx].v=v;
  18     e[idx].w=w;
  19     e[idx].ne=head[u];
  20     head[u]=idx;
  21 }
  22  
  23 void build(){
  24     cin>>n>>m;
  25     int a,b;   ll c;
  26     for (int i=0; i<m; i++) {
  27         scanf("%d%d%lld",&a,&b,&c);
  28         addedge(a,b,c);
  29         addedge(b,a,c);
  30     }
  31 }
  32  
  33 void dfs(int u,int cnt){
  34     d[u]=cnt;
  35     for (int i=head[u]; i; i=e[i].ne){
  36         int v=e[i].v;  ll w=e[i].w;
  37         if(d[v]>d[u]) continue;
   38 is          IF (D [V]) {
   39              LL X = tmp [V] ^ tmp [U] ^ W;   // This value is xor ring
   40              // COUT << "now U =" U << << " = V "V << <<" and X = "X << << endl; 
  41 is              IF (X) A [++ IDXa] = X;   // 0 can discard 
  42 is          } the else {
   43 is              tmp [V] tmp = [U] ^ W;   // before dfs (u), u has a value calculated tmp 
  44 is              DFS (V, + CNT . 1 );
  45         }
  46     }
  47 }
  48  
  49 void xxj(){
  50     for (int i=1; i<=idxa; i++) for (int j=63; j>=0; j--) if((1LL<<j)&a[i]){
  51         if(b[j]) a[i]^=b[j];
  52         else { b[j]=a[i]; break; }
  53     }
  54     ll ans=tmp[n];
  55     for (int i=63; i>=0; i--) ans=max(ans,ans^b[i]);
  56     printf("%lld\n",ans);
  57 }
  58  
  59 int main(){
  60     build();
  61     dfs(1,1);
  62     xxj();
  63     return 0;
  64 }

  The word