POJ 2186 -- Popular Cows (Strongly Connected Components)

Problem: http://poj.org/problem?id=2186

缩点 http://en.wikipedia.org/wiki/Vertex_contraction#Vertex_identification

强连通分量

http://en.wikipedia.org/wiki/Strongly_connected_component

http://en.wikipedia.org/wiki/Kosaraju%27s_algorithm

http://en.wikipedia.org/wiki/Tarjan%27s_strongly_connected_components_algorithm

http://en.wikipedia.org/wiki/Path-based_strong_component_algorithm

Popular Cows

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 22027 Accepted: 8993

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 31 22 12 3

Sample Output

1

Hint

Cow 3 is the only cow of high popularity. 

Source

USACO 2003 Fall

Kosaraju's algorithm works as follows:

• Let G be a directed graph and S be an empty stack.
• While S does not contain all vertices:
  • Choose an arbitrary vertex v not in S. Perform a depth-first search starting at v. Each time that depth-first search finishes expanding a vertex u, push u onto S.
• Reverse the directions of all arcs to obtain the transpose graph.
• While S is nonempty:
  • Pop the top vertex v from S. Perform a depth-first search starting at v in the transpose graph. The set of visited vertices will give the strongly connected component containing v; record this and remove all these vertices from the graph G and the stack S. Equivalently, breadth-first search (BFS) can be used instead of depth-first search.

C++ Code:

#include <cstdio>#include <set>#include <stack>#include <vector>using namespace std;void dfs(const int u, const vector<vector<int> > &adj, vector<bool> &flg, stack<int> &stk) {flg[u] = true;const int k = adj[u].size();for (int i = 0; i < k; ++i) {const int v = adj[u][i];if (!flg[v]) dfs(v, adj, flg, stk);}stk.push(u);}void scc(const int u, const int cnt, const vector<vector<int> > &rev, vector<bool> &flg, vector<int> &color) {flg[u] = true;color[u] = cnt;const int k = rev[u].size();for (int i = 0; i < k; ++i) {const int v = rev[u][i];if (!flg[v]) scc(v, cnt, rev, flg, color);}}int main() {int n, m;scanf("%d%d", &n, &m);vector<vector<int> > adj(n, vector<int>());vector<vector<int> > rev(n, vector<int>());while (m--) {int a, b;scanf("%d%d", &a, &b);--a, --b;adj[a].push_back(b);rev[b].push_back(a);}stack<int> stk;vector<bool> flg(n, false);for (int i = 0; i < n; ++i) {if (flg[i]) continue;dfs(i, adj, flg, stk);}flg.clear();flg.resize(n, false);vector<int> color(n, -1);int cnt = 0;while (!stk.empty()) {const int u = stk.top();stk.pop();if (flg[u]) continue;scc(u, cnt, rev, flg, color);++cnt;}vector<int> num(cnt, 0);for (int i = 0; i < n; ++i)++num[color[i]];vector<set<int> > edge(cnt, set<int>());for (int i = 0; i < n; ++i) {const int k = adj[i].size();for (int j = 0; j < k; ++j)if (color[i] != color[adj[i][j]])edge[color[i]].insert(color[adj[i][j]]);}int ans = 0;for (int i = 0; i < cnt; ++i) {if (edge[i].empty()) {if (ans == 0)ans = num[i];else {ans = 0;break;}}}printf("%d\n", ans);return 0;}