uva 11504 - Dominos（强联通分量）

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题目链接：uva 11504 - Dominos

缩点，入度为0的点个数即为答案。

#include <cstdio>#include <cstring>#include <vector>#include <stack>#include <algorithm>using namespace std;const int maxn = 1e5 + 5;int N, M, in[maxn];int cntlock, cntscc, pre[maxn], sccno[maxn], low[maxn];vector<int> G[maxn];stack<int> S;void dfs (int u) {pre[u] = low[u] = ++cntlock;S.push(u);for (int i = 0; i < G[u].size(); i++) {int v = G[u][i];if (!pre[v]) {dfs(v);low[u] = min(low[u], low[v]);} else if (!sccno[v])low[u] = min(low[u], pre[v]);}if (low[u] == pre[u]) {cntscc++;while (true) {int x = S.top(); S.pop();sccno[x] = cntscc;if (x == u) break;}}}void findSCC() {cntlock = cntscc = 0;memset(pre, 0, sizeof(pre));memset(sccno, 0, sizeof(sccno));for (int i = 1; i <= N; i++)if (!pre[i]) dfs(i);}void init () {scanf("%d%d", &N, &M);for (int i = 1; i <= N; i++) G[i].clear();int u, v;while (M--) {scanf("%d%d", &u, &v);G[u].push_back(v);}findSCC();}int solve () {memset(in, 0, sizeof(in));for (int i = 1; i <= N; i++) {int u = sccno[i];for (int j = 0; j < G[i].size(); j++) {int v = sccno[G[i][j]];if (u != v) in[v] = 1;}}int ret = 0;for (int i = 1; i <= cntscc; i++)if (!in[i]) ret++;return ret;}int main () {int cas;scanf("%d", &cas);while (cas--) {init();printf("%d\n", solve());}return 0;}

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