POJ 2186 —— 分解强连通分量
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Popular Cows
Description
Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that cow B is popular. Since popularity is transitive, if A thinks B is popular and B thinks C is popular, then A will also think that C is
popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow. Input
* Line 1: Two space-separated integers, N and M
* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular. Output
* Line 1: A single integer that is the number of cows who are considered popular by every other cow.
Sample Input 3 3 1 2 2 1 2 3 Sample Output 1 Hint
Cow 3 is the only cow of high popularity.
Source
USACO 2003 Fall
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题意是有n头牛，m个关系，代表牛A认为牛B是明星，关系具有传递性：A->B 且B->C 则A->C。输出被所有牛都认为是明星的牛数。下面为了方便把这种牛称为超级牛。
思路：在一个图的SCC中如果有一头超级牛，则这个SCC中所有牛都是超级牛。分解SCC后，至多有一个SCC满足题意，就是拓扑序最后的SCC，然后我们去检查这个SCC是否是所有顶点都可以到达的，rdfs一次即可，最后输出该SCC的节点数。
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAX_V = 10000 + 50;
const int MAXS = 50000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
const int inf = 1 << 30;
#define eps 1e-10
const long long MOD = 1000000000 + 7;
const int mod = 10007;
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pii;
typedef vector<int> vec;
typedef vector<vec> mat;
#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000")
int V;///顶点数
vector<int> G[MAX_V];///图的邻接表表示
vector<int> rG[MAX_V];///把边反向之后的图
vector<int> vs;///后序遍历顺序的顶点列表
bool used[MAX_V];///访问标记
int cmp[MAX_V];///所属强联通分量的拓扑序
void add_edge(int from , int to)
{
G[from].push_back(to);
rG[to].push_back(from);
}
void dfs(int v)
{
used[v] = true;
for(int i = 0; i < G[v].size() ; i++)
{
if(!used[G[v][i]])dfs(G[v][i]);
}
vs.push_back(v);
}
void rdfs(int v , int k)
{
used[v] = true;
cmp[v] = k;
for(int i = 0 ; i < rG[v].size() ; i++)
{
if(!used[rG[v][i]])rdfs(rG[v][i] , k);
}
}
int scc()
{
memset(used , 0 , sizeof(used));
vs.clear();
for(int v = 0 ; v < V ; v++)
{
if(!used[v])dfs(v);
}
memset(used , 0 , sizeof(used));
int k = 0;
for(int i = vs.size() - 1 ; i >= 0 ; i--)
{
if(!used[vs[i]])rdfs(vs[i] , k++);
}
return k;
}
int n , m;
int a[MAXS] , b[MAXS];
void solve()
{
V = n;
for(int i = 0 ; i < m ; i ++)add_edge(a[i] - 1 , b[i] - 1);
int numofscc = scc();
int u = 0 , num = 0;
for(int v = 0 ; v < V ; v++)
{
if(cmp[v] == numofscc - 1)
{
u = v;
num++;
}
}
memset(used ,0 , sizeof(used));
rdfs(u , 0);
for(int v = 0 ; v < V ; v++)
{
if(!used[v])
{
num = 0;
break;
}
}
printf("%d\n" , num);
}
int main()
{
while(~scanf("%d%d" , &n , &m))
{
for(int i = 0 ; i < m ; i++)
{
scanf("%d%d" , &a[i] , &b[i]);
}
solve();
}
return 0;
}