bzoj3346:ural1811 Dual Sim Phone

First, the side to go heavy, then there is $n\geq\sqrt{m}$.

Then the two-point answer, converted to determine whether there are two points of their out-side set of sets for the complete set.

Then these two points must satisfy $deg_x+deg_y\geq n$.

$deg_x\geq deg_y$ may be set, then there is $deg_x\times 2\geq n$.

Consider enumerating $x$, with a maximum of $o (\frac{m}{n}) $ $x$.

To enumerate $y$ again, there are two kinds of decision algorithms:

$1.$ $f[i][j]$ Indicates whether $i$ is not pointing to $j$, so long as there is $f[x][j]\ and\ f[y][j]=true$ is not feasible.

can be used to calculate the time complexity of $o (\frac{nm}{32}) $.

$2.$ enumerates all the out edges of the $y$ and determines whether or not it appears in $x$ by a timestamp.

Time Complexity $o (\frac{m^2}{n}) $.

Set $s$ as the threshold value, when $n\leq S $ is used algorithm 1, otherwise, the algorithm 2, there is $\frac{sm}{32}\leq\frac{m^2}{s}$.

When $s$ takes $\sqrt{\frac{m}{32}}$, it obtains the optimal complexity of $o (M\sqrt{\frac{m}{32}}) $.

Total time complexity $o (M\log m\sqrt{\frac{m}{32}}) $.

#include <cstdio> #include <algorithm>const int n=10005,m=100010;int n,m,u,i,j,k,a[m],st[n],en[n],d[n],v [n],t,l,r,mid,ans;unsigned int f[2048][64];struct e{int x, Y, Z;}  e[m];inline BOOL CMP (const E&a,const e&b) {if (a.x!=b.x) return a.x<b.x;  if (A.Y!=B.Y) return a.y<b.y; return a.z<b.z;} inline void Read (int&a) {char c;while (!) ( ((C=getchar ()) >= ' 0 ') && (c<= ' 9 ')); a=c-' 0 '; while (((C=getchar ()) >= ' 0 ') && (c<= ' 9 ')) (a*= Ten) +=c-' 0 ';}      BOOL Check (int mid) {if (n>2048) {for (i=0;i<n;i++) {d[i]=0;    for (j=st[i];j<en[i];j++) if (e[j].z<=mid) d[i]++;    } for (i=0;i<n;i++) v[i]=-1;      for (i=0;i<n;i++) if (d[i]*2>=n) {for (k=st[i];k<en[i];k++) if (e[k].z<=mid) v[e[k].y]=i;        for (j=0;j<n;j++) if (D[i]+d[j]>=n&&d[i]>=d[j]) {t=d[i];        for (k=st[j];k<en[j];k++) if (e[k].z<=mid&&v[e[k].y]<i) t++;      if (t==n) return 1; }}}else{for (i=0;i<n;i++) {for (j=0;j<u;J + +) f[i][j]=~0u;      f[i][u]=0;    for (j=u<<5;j<n;j++) f[i][j>>5]|=1u<< (j&31);      } for (i=0;i<n;i++) {d[i]=0;    for (j=st[i];j<en[i];j++) if (e[j].z<=mid) d[i]++,f[i][e[j].y>>5]^=1u<< (e[j].y&31);        } for (i=0;i<n;i++) if (d[i]*2>=n) {for (j=0;j<n;j++) if (D[i]+d[j]>=n&&d[i]>=d[j]) {t=1;        for (k=0;k<=u;k++) if (F[i][k]&f[j][k]) {t=0;break;}      if (t) return 1; }}} return 0;} int main () {read (n), read (m);  u= (n-1) >>5;    for (i=1;i<=m;i++) {read (e[i].x), read (E[I].Y), read (E[I].Z);  e[i].x--, e[i].y--;  } std::sort (E+1,E+M+1,CMP); for (i=1;i<=m;i++) if (i==1| | e[i].x!=e[i-1].x| |  E[I].Y!=E[I-1].Y) E[++j]=e[i];  for (m=j,i=1;i<=m;i++) a[i]=e[i].z;  Std::sort (a+1,a+m+1);    for (i=0,j=1;i<n;i++) {st[i]=j;    while (j<=m&&e[j].x==i) j + +;  En[i]=j;  } l=1,r=m;  while (l<=r) if (check (a[mid= (l+r) >>1]) r= (ans=mid) -1;else l=mid+1; if (!ans) puts ("No solution"); else printf ("%d ", A[ans]); return 0;}

　　

bzoj3346:ural1811 Dual Sim Phone