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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1269
思路分析:该问题要求判断是否每两个房间都可以相互到达,即求该有向图中的所有点是否只构成一个强连通图分量,使用Tarjan算法即可求解;
代码如下:
#include <stack> #include <vector> #include <cstdio> #include <cstring> #include <iostream> using namespace std; const int MAX_N = 10000 + 100; vector<int> G[MAX_N]; stack<int> S; int pre[MAX_N], lowlink[MAX_N], scc_no[MAX_N]; int dfs_clock, scc_cnt; inline int Min(int a, int b) { return a < b ? a : b; } void Dfs(int u) { pre[u] = lowlink[u] = ++dfs_clock; S.push(u); for (int i = 0; i < G[u].size(); ++i) { int v = G[u][i]; if (!pre[v]) { Dfs(v); lowlink[u] = Min(lowlink[u], lowlink[v]); } else if (!scc_no[v]) lowlink[u] = Min(lowlink[u], lowlink[v]); } if (lowlink[u] == pre[u]) { scc_cnt++; for (;;) { int x = S.top(); S.pop(); scc_no[x] = scc_cnt; if (x == u) break; } } } inline void FindScc(int n) { dfs_clock = scc_cnt = 0; memset(pre, 0, sizeof(pre)); memset(lowlink, 0, sizeof(lowlink)); memset(scc_no, 0, sizeof(scc_no)); for (int i = 1; i <= n; ++i) if (!pre[i]) Dfs(i); } int main() { int ver_num, road_num; while (scanf("%d %d", &ver_num, &road_num) != EOF && (ver_num + road_num)) { int ver_1, ver_2; for (int i = 0; i < MAX_N; ++i) G[i].clear( ); for (int i = 0; i < road_num; ++i) { scanf("%d %d", &ver_1, &ver_2); G[ver_1].push_back(ver_2); } FindScc(ver_num); while (!S.empty( )) S.pop( ); if (scc_cnt > 1) printf("No\n"); else printf("Yes\n"); } return 0; }
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原文地址:http://www.cnblogs.com/tallisHe/p/4674970.html