链接:https://vjudge.net/problem/HDU-2767
题意:
给一个图,求最少需要几条边将其连成一个强连通图
思路:
tarjan,缩点,考虑缩点后的图,出度为0的点和入度为0的点,而所需要的边就是出度为0,和入度为0的点的较大值。
代码:
#include <iostream> #include <memory.h> #include <string> #include <istream> #include <sstream> #include <vector> #include <stack> #include <algorithm> #include <map> #include <queue> #include <math.h> #include <cstdio> #include <set> #include <iterator> #include <cstring> using namespace std; typedef long long LL; const int MAXN = 2e4+10; vector<int> G[MAXN]; stack<int> St; int Dfn[MAXN], Low[MAXN]; int Vis[MAXN], Dis[MAXN][2]; // 0:in, 1:out int Fa[MAXN]; int n, m; int times, cnt; void Init() { for (int i = 1;i <= n;i++) G[i].clear(), Fa[i] = i; memset(Dfn, 0, sizeof(Dfn)); memset(Low, 0, sizeof(Low)); memset(Vis, 0, sizeof(Vis)); memset(Dis, 0, sizeof(Dis)); times = cnt = 0; } void Tarjan(int x) { St.push(x); Vis[x] = 1; Dfn[x] = Low[x] = ++times; for (int i = 0;i < G[x].size();i++) { int node = G[x][i]; if (Dfn[node] == 0) { Tarjan(node); Low[x] = min(Low[x], Low[node]); } else if (Vis[node] == 1) Low[x] = min(Low[x], Dfn[node]); } if (Low[x] == Dfn[x]) { ++cnt; while (x != St.top()) { Fa[St.top()] = cnt; Vis[St.top()] = 0; St.pop(); } Fa[St.top()] = cnt; Vis[St.top()] = 0; St.pop(); } } int main() { int t; cin >> t; while (t--) { cin >> n >> m; Init(); int l, r; for (int i = 1;i <= m;i++) { cin >> l >> r; G[l].push_back(r); } for (int i = 1;i <= n;++i) if (!Dfn[i]) Tarjan(i); if (cnt == 1) cout << 0 << endl; else { for (int i = 1;i <= n;i++) { for (int j = 0;j < G[i].size();j++) { int node = G[i][j]; if (Fa[i] != Fa[node]) ++Dis[Fa[i]][1], ++Dis[Fa[node]][0]; } } int in = 0, out = 0; for (int i = 1;i <= cnt;i++) { if (Dis[i][0] == 0) in++; if (Dis[i][1] == 0) out++; } cout << max(out, in) << endl; } } return 0; }