UVA 11504 - Dominos(强连通分量)

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UVA 11504 - Dominos

题目链接

题意：给定一个多米诺骨牌的有向图，为最多要推几个才能全倒

思路：强连通分量，缩点后找出度数为0的点就是答案

代码：

#include <cstdio>#include <cstring>#include <vector>#include <stack>#include <algorithm>using namespace std;const int N = 100005;vector<int> g[N], scc[N];int pre[N], lowlink[N], sccno[N], dfs_clock, scc_cnt;stack<int> S;void dfs_scc(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_scc(v);lowlink[u] = min(lowlink[u], lowlink[v]);} else if (!sccno[v])lowlink[u] = min(lowlink[u], pre[v]);}if (lowlink[u] == pre[u]) {scc_cnt++;while (1) {int x = S.top(); S.pop();sccno[x] = scc_cnt;if (x == u) break;}}}void find_scc(int n) {dfs_clock = scc_cnt = 0;memset(sccno, 0, sizeof(sccno));memset(pre, 0, sizeof(pre));for (int i = 0; i < n; i++)if (!pre[i]) dfs_scc(i);}int in[N];int t, n, m, u[N], v[N];int main() {scanf("%d", &t);while (t--) {scanf("%d%d", &n, &m);for (int i = 0; i < n; i++)g[i].clear();for (int i = 0; i < m; i++) {scanf("%d%d", &u[i], &v[i]);u[i]--; v[i]--;g[u[i]].push_back(v[i]);}find_scc(n);memset(in, 0, sizeof(in));for (int i = 0; i < m; i++) {if (sccno[u[i]] != sccno[v[i]])in[sccno[v[i]]]++;}int ans = 0;for (int i = 1; i <= scc_cnt; i++)if (in[i] == 0) ans++;printf("%d\n", ans);}return 0;}

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