POJ 2186 Popular Cows (kosaraju+ strong Unicom component template)

Title Link: http://poj.org/problem?id=2186

The main topic: given n cattle and M ordered pairs (A, B), (A, B) said that A Bull is a red man, the relationship has transitivity, if cow a think that cow B is a red, cow B is the bull C is the red, then cow A also think that the Ox C is the Reds. The total number of cows considered by all other cows to be Red Bull.

The idea of solving a problem: to treat all cows as vertices and order pairs (a, a, A, b) as a forward edge from a to a, the title becomes the total number of vertices that all vertices can reach. We can get a conclusion that if a strong connected component has a cow that is considered a red, then all the cows in the strong Unicom component are red, which is clearly correct. Since I use Kosaraju to find strong unicom components, according to the algorithm nature, Red Bull will only be in the final topological sequence of strong unicom components, I only need to locate the last piece of strong unicom components, take one of the vertices, to see if all the points can be reached.

1#include <iostream>2#include <cstring>3#include <algorithm>4#include <vector>5 using namespacestd;6 Const intn=1e4+5;7 8vector<int>G[N];//adjacency Table of graphs9vector<int>rG[N];//adjacency table for reverse graphsTenvector<int>vs;//list of vertices of sequential traversal order One BOOLUsed[n];//whether the record point is accessed A intCmp[n];//Cmp[i] represents the topological sequence of the strong unicom component of point I -  - intv,e; the  - voidAddedge (intUintv) { - G[u].push_back (v); - rg[v].push_back (u); + }  -  + voidDfsintv) { Aused[v]=true; at      for(intI=0; I<g[v].size (); i++){ -         if(!Used[g[v][i]]) - DFS (G[v][i]); -     } -     //marking prior to backtracking - Vs.push_back (v); in } -  to voidRdfsintVintk) { +used[v]=true; -     //Point v belongs to the K-strong connected component thecmp[v]=K; *      for(intI=0; I<rg[v].size (); i++){ $         if(!Used[rg[v][i]])Panax Notoginseng Rdfs (rg[v][i],k); -     } the } +  A intSCC () { thememset (Used,false,sizeof(used)); + vs.clear (); -     //First time Dfs $      for(intI=1; i<=v;i++){ $         if(!Used[i]) - DFS (i); -     } thememset (Used,false,sizeof(used)); -     intk=0;//number of strong Unicom component blocksWuyi     //Dfs for the second time the      for(intI=vs.size ()-1; i>=0; i--){ -         if(!Used[vs[i]]) WuRdfs (vs[i],++k); -     } About     returnK; $ } -  - voidsolve () { -     //get the number of strong unicom blocks A     intn=SCC (); +     //number of statistical alternative solutions the     intu=0, num=0; -      for(intI=1; i<=v;i++){ $         if(cmp[i]==N) { theu=i; thenum++; the         } the     } -     //Check if all points are up to inmemset (Used,0,sizeof(used)); theRdfs (U,0); the      About      for(intI=1; i<=v;i++){ the         if(!Used[i]) { thenum=0; the              Break; +         } -     } the      for(intI=1; i<=v;i++){Bayicout<<"i="<<i<<"cmp="<<cmp[i]<<Endl; the     } theprintf"%d\n", num); - } -  the intMain () { thescanf"%d%d",&v,&E); the      for(intI=1; i<=e;i++){ the         intu,v; -scanf"%d%d",&u,&v); the Addedge (u,v); the     } the solve ();94     return 0;  the}

POJ 2186 Popular Cows (kosaraju+ strong Unicom component template)