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题目大意:有n个牛在一块, m条单项绳子, 有m个链接关系, 问有多少个团体内部任意两头牛可以相互可达
解题思路:有向图强连通分量模版图
代码如下:
#include<stdio.h> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; typedef long long ll; const int N = 10003; vector<int>G[N], DQ; int low[N], dfn[N], tot; bool mk[N]; int n, m, ans; void init() { ans = tot = 0; DQ.clear(); for(int i=1; i<=n; ++ i) { G[i].clear(); low[i] = dfn[i] = -1; mk[i] = false; } } void tarjan(int u, int f) { dfn[u] = low[u] = ++ tot; DQ.push_back(u); mk[u] = true; for(int i = 0; i<G[u].size(); ++ i) { int v = G[u][i]; if(dfn[v] == -1) { tarjan(v, u); low[u] = min(low[u], low[v]); } else if(mk[v]) low[u] = min(low[u], dfn[v]); } if(dfn[u] == low[u]) { int s; int k = 0; do { s = DQ.back(); k ++; DQ.pop_back(); mk[s] = false; } while(u != s); if(k > 1) ans ++; } } void solve() { for(int i=1; i<=n; ++ i) { if(dfn[i] == -1) tarjan(i, -1); } printf("%d\n", ans); } int main() { while(~scanf("%d%d", &n, &m)) { init(); for(int i=1; i<=m; ++ i) { int u, v; scanf("%d%d", &u, &v); G[u].push_back(v); } solve(); } return 0; }
POJ 3180-The Cow Prom (图论-有向图强联通tarjan算法)
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原文地址:http://www.cnblogs.com/aiterator/p/5935634.html