POJ 1236 Network of Schools(强连通分量)
POJ 1236 Network of Schools
题意:题意本质上就是,给定一个有向图,问两个问题
1、从哪几个顶点出发,能走全所有点
2、最少连几条边,使得图强连通
思路:
#include <cstdio> #include <cstring> #include <vector> #include <stack> using namespace std; const int N = 105; int n; vector<int> g[N]; int pre[N], sccno[N], dfn[N], dfs_clock, sccn; stack<int> S; void dfs_scc(int u) { pre[u] = dfn[u] = ++dfs_clock; S.push(u); for (int i = 0; i < g[u].size(); i++) { int v = g[u][i]; if (!pre[v]) { dfs_scc(v); dfn[u] = min(dfn[u], dfn[v]); } else if (!sccno[v]) dfn[u] = min(dfn[u], pre[v]); } if (dfn[u] == pre[u]) { sccn++; while (1) { int x = S.top(); S.pop(); sccno[x] = sccn; if (x == u) break; } } } void find_scc() { sccn = dfs_clock = 0; memset(pre, 0, sizeof(pre)); memset(sccno, 0, sizeof(sccno)); for (int i = 1; i <= n; i++) if (!pre[i]) dfs_scc(i); } int in[N], out[N]; int main() { while (~scanf("%d", &n)) { int v; for (int i = 1; i <= n; i++) g[i].clear(); int cnt = n; for (int u = 1; u <= n; u++) { while (~scanf("%d", &v) && v) g[u].push_back(v); } find_scc(); memset(in, 0, sizeof(in)); memset(out, 0, sizeof(out)); for (int u = 1; u <= n; u++) { for (int j = 0; j < g[u].size(); j++) { int v = g[u][j]; if (sccno[u] != sccno[v]) { in[sccno[v]]++; out[sccno[u]]++; } } } int ins = 0, outs = 0; for (int i = 1; i <= sccn; i++) { if (!in[i]) ins++; if (!out[i]) outs++; } int ans1 = ins, ans2 = max(ins, outs); if (sccn == 1) ans2 = 0; printf("%d\n%d\n", ans1, ans2); } return 0; }
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