题意: 公司里面的人际关系,A相信B,B相信C,即A相信C,每个人都相信自己,求出存在几个小组,小组内的成员互相相信。
思路:求有向图中,强连通分量的个数。
思路:
#include <iostream> #include <cstdio> #include <cstring> #include <map> #include <algorithm> using namespace std; const int MAXN = 1010; const int MAXM = 999999; struct Edge{ int to, next; }edge[MAXM]; int head[MAXN], tot; int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN]; int Index, top; int scc; bool Instack[MAXN]; int num[MAXN]; int n, m, cnt; map<string, int> name; void init() { tot = cnt = 0; memset(head, -1, sizeof(head)); name.clear(); } void addedge(int u, int v) { edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } void Tarjan(int u) { int v; Low[u] = DFN[u] = ++Index; Stack[top++] = u; Instack[u] = true; for (int i = head[u]; i != -1; i = edge[i].next) { v = edge[i].to; 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; num[scc]++; } while (v != u); } } void solve() { memset(Low, 0, sizeof(Low)); memset(DFN, 0, sizeof(DFN)); memset(num, 0, sizeof(num)); memset(Stack, 0, sizeof(Stack)); memset(Instack, false, sizeof(Instack)); Index = scc = top = 0; for (int i = 1; i <= n; i++) if (!DFN[i]) Tarjan(i); } int main() { while (scanf("%d%d", &n, &m)) { getchar(); if (n == 0 && m == 0) break; init(); string s, s1; for (int i = 0; i < n; i++) { cnt++; getline(cin, s); name[s] = cnt; } for (int i = 0; i < m; i++) { getline(cin, s); getline(cin, s1); addedge(name[s], name[s1]); } solve(); printf("%d\n", scc); } return 0; }
时间: 2024-10-06 20:22:33