[CCF CSP201709-4通信网络]缩点+拓扑+bitset

分类：Data Structure strongly connected components topology bitset

1. 题目链接

[CCF CSP201709-4通信网络]缩点+拓扑

2. 题意描述

3. 解题思路

先对有向图缩点，得到DAG图，和反向DAG图。 然后用bitset维护出每个节点的自身以及他的祖先，对DAG，反向DAG图跑一个拓扑排序，在拓扑排序的时候更新bitset。 然后正向DAG与反向DAG跑出来的bitset做遍或运算。就可以表示出从当前节点出发能够到达的点以及能够达到当前节点的所以顶点。 不过，好像代码写得有点挫，另外，csp不支持c++11，考试的时候千万别用c++11啦，千万别偷懒，不然就爆0了。

4. 实现代码

#include <bits/stdc++.h> using namespace std; typedef long long ll; typedef pair<int, int> pii; typedef pair<ll, ll> pll; const int inf = 0x3f3f3f3f; const ll infl = 0x3f3f3f3f3f3f3f3fLL; template<typename T> inline void umax(T &a, T b) { a = max(a, b); } template<typename T> inline void umin(T &a, T b) { a = min(a, b); } const int MAXN = 1005; vector<int> T[MAXN], G[2][MAXN]; bitset<MAXN> bs[2][MAXN], okay; int n, m; int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN]; //Belong 数组的值是 1~scc int Index, top; int scc;//强连通分量的个数 bool Instack[MAXN]; void Tarjan(int u) { int v; Low[u] = DFN[u] = ++Index; Stack[top++] = u; Instack[u] = true; for (int i = 0; i < T[u].size(); ++i) { int v = T[u][i]; if (!DFN[v]) { Tarjan(v); if (Low[u] > Low[v]) Low[u] = Low[v]; } else if (Instack[v] && Low[u] > DFN[v]) Low[u] = DFN[v]; } if (Low[u] == DFN[u]) { scc++; do { v = Stack[--top]; Instack[v] = false; Belong[v] = scc; } while (v != u); } } int du[2][MAXN]; void topu(int f) { for (int i = 1; i <= scc; ++i) { for (int e = 0; e < G[f][i].size(); ++e) { int j = G[f][i][e]; ++ du[f][j]; } } queue<int> q; for (int i = 1; i <= scc; ++i) { bs[f][i][i] = 1; if (!du[f][i]) q.push(i); } while (!q.empty()) { int u = q.front(); q.pop(); for (int i = 0; i < G[f][u].size(); ++i) { int v = G[f][u][i]; -- du[f][v]; bs[f][v] |= bs[f][u]; if (du[f][v] != 0) continue; q.push(v); } } } int main() { #ifdef ___LOCAL_WONZY___ freopen("input.txt", "r", stdin); #endif // ___LOCAL_WONZY___ int u, v; cin >> n >> m; for (int i = 1; i <= m; ++i) { cin >> u >> v; // G[0][u].push_back(v); // G[1][v].push_back(u); T[u].push_back(v); } Index = scc = top = 0; for (int i = 1; i <= n; ++i) { if (!DFN[i]) Tarjan(i); } for (int i = 1; i <= n; ++i) { for (int e = 0; e < T[i].size(); ++e) { int j = T[i][e]; if (Belong[i] == Belong[j]) continue; G[0][Belong[i]].push_back(Belong[j]); G[1][Belong[j]].push_back(Belong[i]); } } for (int i = 1; i <= scc; ++i) { sort(G[0][i].begin(), G[0][i].end()); G[0][i].erase(unique(G[0][i].begin(), G[0][i].end()), G[0][i].end()); sort(G[1][i].begin(), G[1][i].end()); G[1][i].erase(unique(G[1][i].begin(), G[1][i].end()), G[1][i].end()); } topu(0); topu(1); for (int i = 1; i <= scc; ++i) { if (okay[i]) continue; bs[0][i] |= bs[1][i]; int a = bs[0][i].count(); if (a == scc) okay[i] = true; } int ans = 0; for (int i = 1; i <= n; ++i) { if (okay[Belong[i]]) ++ ans; } printf("%d\n", ans); #ifdef ___LOCAL_WONZY___ cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC * 1000 << "ms." << endl; #endif // ___LOCAL_WONZY___ return 0; }