CF555E Case of Computer Network

Portal Test Instructions

Exercises

Easy to find, in the same side of the double inside the point we do not care about it, it is sure to.

So let's pinch and turn the graph into a tree. This way, the path of $s $ to $t $ can be $\mathrm{lca}$.

We put the $s $ to $\mathrm{lca}$ the path up the marker, put the $\mathrm{lca}$ to the $t $ path to hit the downward marker.

As long as there are no paths with two kinds of markers, it is possible.

Marking uses a differential implementation on the tree. Time complexity: $\mathcal o\left (n+m+q\mathrm{lg}n\right) $.

Attached code:

#pragmaGCC optimize (3)#include<bits/stdc++.h>using namespacestd;#defineDown (x, y) x = min (x, y)intN, M, q;structEdge {int  from, to, NXT, id;} edges[400005];structEdgE {intto, NXT;} edges[400005];intE0, E =1, e2 =1, hed[200005], hed[200005], Tim =0, fa[200005], dfn[200005], low[200005], scc[200005], dis[200005], lca[ -][200005];intflg[200005];BOOLroot[200005];BOOLgoon[200005];intleaves[200005], top=0;Const intClean_flag =0x1;Const intUp_flag =0x2;Const intDown_flag =0x4; Stack<int>S;inlineBOOLHasflag (intXintattr) {    returnFLG[X] &attr;} InlineBOOLSetflag (intXintattr) {    if(attr = = Up_flag && hasflag (x, Down_flag))return false; if(attr = = Down_flag && hasflag (x, Up_flag))return false; FLG[X]|=attr; return true;} InlinevoidAddedge (intXinty) {edges[e]=(Edge) {x, Y, hed[x], E2}; HED[X]= e++; Edges[e]= (edge) {y, X, hed[y], e2++}; Hed[y]= e++;} InlinevoidAddedge (intXinty) {edges[e]=(EdgE) {y, hed[x]}; HED[X]= e++;} Inlinevoid_tarjan (intx) {dfn[x]= Low[x] = + +Tim;    S.push (x);  for(intE=HED[X]; E E=edges[e].nxt) {        inty =edges[e].to; if(Edges[e].id = = Fa[x])Continue; if(!Dfn[y]) {Fa[y]=edges[e].id;            _tarjan (y);        Down (Low[x], low[y]); } ElseDown (Low[x], dfn[y]); }    if(Dfn[x] = =Low[x]) {Scc[x]= ++E2;  while(S.top ()! =x) {scc[s.top ()]=E2;        S.pop ();    } s.pop (); }}inlinevoidTarjan () { for(intI=1; i<=n; i++)if(!Dfn[i]) _tarjan (i);  for(inti =0; i < E0; i++) {        intx = Edges[i]. from, y =edges[i].to; if(Scc[x] = = Scc[y])Continue;    Addedge (Scc[x], scc[y]); }}inlinevoidLca_pre () { for(intI=1; i< -; i++)         for(intj=1; j<=n; J + +) Lca[i][j]= lca[i-1][lca[i-1][j]];} InlineintLCA (intXinty) {if(Dis[x] >Dis[y]) swap (x, y); intU = dis[x], V =Dis[y]; intX=x, y=y;  for(intDet = v-u, i=0; Det det>>=1, ++i)if(Det &1* Z =Lca[i][y]; if(X = = Y)returnX;  for(intI= -; i>=0; i--) {        if(Lca[i][x] = = Lca[i][y])Continue; X= Lca[i][x]; Y =Lca[i][y]; }    returnlca[0][x];}intgroup[200005], Current;inlinevoidDfsintXintFA) {    BOOLIsLeaf =true; GROUP[X] =Current ;  for(intE=HED[X]; E E=edges[e].nxt) {        inty =edges[e].to; if(Y! =FA) {IsLeaf=false; lca[0][y] =x; Dis[y]= Dis[x] +1;        DFS (y, x); }    }    if(isleaf) leaves[top++] =x;} InlineBOOLAkmachine (intXintflag) {     while(!Root[x]) {        if(Goon[x])return true; GOON[X]=true; if(Hasflag (x, Up_flag) &&!hasflag (x, Clean_flag) && (FLAG &Down_flag)|| Hasflag (x, Down_flag) &&!hasflag (x, Clean_flag) && (FLAG &Up_flag)) return false; if(Hasflag (x, clean_flag)) FLAG =0; Flag= Flg[x] & (Up_flag |Down_flag); X= lca[0][x]; }    return true;} InlineBOOLCheck () { for(intI=0; i<top; i++)if(!akmachine (Leaves[i),0))return false; return true;} InlineBOOLsolve () { for(intI=1; i<=n; i++)        if(!Dis[i]) { Current= i; lca[0][i] = i; Root[i] =1; DFS (i,0);    } lca_pre ();  while(q--) {        intS, t; CIN >> S >>T; S= Scc[s]; t =Scc[t]; if(Group[s]! = Group[t])return false; if(s = = t)Continue; intp =LCA (S, t); if(P! = s)if(!setflag (S, Up_flag))return false; if(P! = t)if(!setflag (t, Down_flag))return false;    Setflag (P, Clean_flag); }    returncheck ();}intMain () {iOS:: Sync_with_stdio (false); Cin.tie (0); CIN>> n >> M >>Q;  for(intI=1; i<=m; i++) {        intx, y; CIN >> x >>y;    Addedge (x, y); } E0= e; E =1; E2 =0; Tarjan (); cout<< (Solve)?"Yes":"No") <<Endl; return 0;}

CF555E Case of Computer Network