POJ 2186 Popular Cows (strong unicom component)

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

Main topic:

The desire of every cow is to become one of the most popular cows. Now there are n cows, give you m-pairs of integers (a, a, b), indicating that cow A is considered to be popular. This relationship is transitive, and if a thinks B is popular, B thinks C is popular, then bull a also thinks that Ox C is popular. Your task is to find out how many cows are considered popular by all cows. Problem Solving Ideas:Suppose there are two cows A and B are all the other cows are considered to be red, then obviously, a by B is considered to be a red, B is also considered a red, that is, there is a and B two vertices of the circle, or a, b belong to a strong unicom component. So if a cow is considered a red-bull by all the other cows, all the cows within its strong unicom component are considered to be reds by all other cows. After we have a strong connectivity component decomposition, at most, there is a strong unicom component to meet the conditions of the problem. Practice: First use Tarjan to find each strong connected components, and then shrink points, statistics of each point out, if there is only 1 out of 0 points, the output of this point contains the number of nodes, otherwise output 0. Code

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5#include <vector>6#include <stack>7 using namespacestd;8 Const intn=1e4+5;9 Ten intCnt,num; One intDfn[n],low[n],fa[n],sze[n],outdeg[n]; Astack<int>SK; -vector<int>V[n]; -  the voidTarjan (intu) { -dfn[u]=low[u]=++CNT; - sk.push (u); -      for(intI=0; I<v[u].size (); i++){ +         intt=V[u][i]; -         if(!dfn[t]) {//Point T is not accessed + Tarjan (t); Alow[u]=min (low[u],low[t]); at         } -         Else if(!fa[t]) low[u]=min (low[u],dfn[t]);//Point T has been accessed and T is still in the stack -     } -     if(low[u]==Dfn[u]) { -num++; -          while(1){ in             intt=sk.top (); - Sk.pop (); toFa[t]=num;//indent operations, and place these points as point Num +sze[num]++; -             if(T==u) Break; the         } *     } $ }Panax Notoginseng  - intMain () { the     intn,m; +scanf"%d%d",&n,&m); A      for(intI=1; i<=m;i++){ the         intb; +scanf"%d%d",&a,&b); - V[a].push_back (b); $     } $      for(intI=1; i<=n;i++){ -         if(!Dfn[i]) Tarjan (i); -     } the      for(intI=1; i<=n;i++){ -          for(intj=0; J<v[i].size (); j + +){Wuyi             intt=V[i][j]; the             if(Fa[t]!=fa[i]) outdeg[fa[i]]++; -         } Wu     } -     //A point with a scale of 0 can only have one after the point is shrunk, otherwise it does not meet the conditional output 0 About     intans=0; $      for(intI=1; i<=num;i++){ -         if(!Outdeg[i]) { -             if(ans>0){ -Puts"0"); A                 return 0; +             } theans=Sze[i]; -         } $     } theprintf"%d\n", ans); the     return 0; the}

POJ 2186 Popular Cows (strong unicom component)