Codeforces 506b/505d Mr Kitayuta ' s technology

Portal:Http://codeforces.com/problemset/problem/506/B

Http://codeforces.com/problemset/problem/505/D

Good question

To n cities, M-Bar has a forward edge, to find the least forward edge so that its composition of the figure and the original image and other potential

For each connected component:

If there is no ring, then only need to n-1 edge to complete the Unicom

If there is a ring, then only the N-side to complete the Unicom

So as long as the connected component is sentenced, then there are several connected components have a ring can

Solution One: undirected graph traverse to find strong connected components and then connect the strong connected components represented by the Unicom Component DFS Award ring, as follows

memory:10440 KB

time:498 MS

1#include <iostream>2#include <algorithm>3#include <Set>4#include <cstdio>5#include <cstdlib>6#include <cmath>7#include <vector>8 using namespacestd;9 #definefor (i,j,k) for (int i=j;i<=k;i++)Ten #defineFORD (i,j,k) for (int i=j;i>=k;i--) One #defineLL Long Long A #defineSZ (x) int (x.size ()) - #defineMAXN 100010 - intn,m,x,y; the BOOLcircle=true; - intVIS[MAXN],VIS2[MAXN]; -vector<int>NEG[MAXN],ADJ[MAXN]; -vector<int>dot; + voidNDFs (intstart) - { + dot.push_back (start); Avis[start]=1; atfor (I,0, SZ (Neg[start])-1) -     { -         if(!Vis[neg[start][i]] NDFs (Neg[start][i]); -          -     } -     return; in } - voidADFsintstart) to { +vis2[start]=1; -for (I,0, SZ (Adj[start])-1) the     { *         if(!Vis2[adj[start][i]]) ADFS (Adj[start][i]);  $         Else if(vis2[adj[start][i]]==1) circle=false; Panax Notoginseng     } -vis2[start]=2; the     return; + } A intMain () the { +Cin>>n>>m; -for (I,1, M) $ { $Cin>>x>>y; - neg[x].push_back (y); - neg[y].push_back (x); the adj[x].push_back (y); - }Wuyi intans=N; thefor (I,1, N) - { Wucircle=1; -     if(!Vis[i]) About     { $ dot.clear (); - NDFs (i);  -for (I,0, SZ (dot)-1) -     if(!Vis2[dot[i]]) ADFS (Dot[i]); Aans-=Circle; +     } the } -cout<<ans<<Endl; $ return 0; the}

I'm wondering if the name of the function is to take AFS or ADFS.

Solution Two: In the undirected graph maintenance and check set to find strong connectivity components and the strong connected components of the Unicom components represented by topological sequencing of the ring, as follows

memory:7820 KB

time:514 MS

1#include <iostream>2#include <algorithm>3#include <Set>4#include <cstdio>5#include <cstdlib>6#include <cmath>7#include <vector>8 using namespacestd;9 #definefor (i,j,k) for (int i=j;i<=k;i++)Ten #defineFORD (i,j,k) for (int i=j;i>=k;i--) One #defineLL Long Long A #defineSZ (x) int (x.size ()) - #defineMAXN 100010 -vector<int>G[MAXN]; the intn,m,x,y,t; - intFATHER[MAXN],VAL[MAXN],CIRCLE[MAXN],VIS[MAXN],CLOCK[MAXN]; - intSetfind (intx) - { +     intHasFather[x]; -     if(fa==x)returnx; +     Else returnfather[x]=Setfind (FA); A } at voidSetunion (intXinty) - { -     intx=setfind (x); -     inty=Setfind (y); -     if(x==y)return; -     if(Val[x]>val[y]) father[y]=X; in     Elsefather[x]=Y; -     if(Val[x]==val[y]) val[x]++; to     return; + } - voidDfsintstart) the { *vis[start]=1; $for (I,0, SZ (G[start])-1)Panax Notoginseng     if(!Vis[g[start][i]]) DFS (g[start][i]); -clock[start]=++T; the     return; + } A intMain () the { +Cin>>n>>m; -for (I,1, N) ${father[i]=i;val[i]=1;} $for (I,1, M) - { -Cin>>x>>y; the g[x].push_back (y); - setunion (x, y);Wuyi } thefor (I,1, N) - if(!Vis[i]) DFS (i); Wufor (I,1, N) -For (J,0, SZ (G[i])-1) About if(Clock[i]<clock[g[i][j]]) circle[setfind (i)]=1; $ intans=N; -for (I,1, N) - if(I==setfind (i))if(!circle[setfind (i)]) ans--; -cout<<ans<<Endl; A return 0; +}

Mr Kitayuta ' s Black technology

In fact, the connectivity component can also be dyed

　　To be able to use and check the set.

Anyway, it's all going to be a mess.

Codeforces 506b/505d Mr Kitayuta ' s technology