uva 1364 - Knights of the Round Table（二分图+强连通）

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题目链接：uva 1364 - Knights of the Round Table

处理出所有点-双联通分量，判断每个联通分量是否是二分图，如果不是二分图的话所有点即为有会议可以参加。

#include <cstdio>#include <cstring>#include <vector>#include <stack>#include <algorithm>using namespace std;typedef pair<int,int> pii;const int maxn = 1005;int N, M, vis[maxn][maxn], color[maxn], odd[maxn];int cntlock, cntbcc, pre[maxn], low[maxn], iscut[maxn], bccno[maxn];stack<pii> S;vector<int> G[maxn], BCC[maxn];bool bipartite (int u, int d) {for (int i = 0; i < G[u].size(); i++) {int v = G[u][i];if (bccno[v] != d) continue;if (color[v] == color[u]) return false;if (!color[v]) {color[v] = 3 - color[u];if (!bipartite(v, d)) return false;}}return true;}int dfs (int u, int fa) {int lowu = pre[u] = ++cntlock, child = 0;for (int i = 0; i < G[u].size(); i++) {int v = G[u][i];pii e = make_pair(u, v);if (!pre[v]) {child++;S.push(e);int lowv = dfs(v, u);lowu = min(lowu, lowv);if (lowv >= pre[u]) {iscut[u] = true;BCC[++cntbcc].clear();while (true) {e = S.top(); S.pop();if (bccno[e.first] != cntbcc) {BCC[cntbcc].push_back(e.first);//bccno[e.first] = cntbcc;}if (bccno[e.second] != cntbcc) {BCC[cntbcc].push_back(e.second);//bccno[e.second] = cntbcc;}if (u == e.first && v == e.second) break;}}} else if (v != fa && pre[v] < pre[u]) {S.push(e);lowu = min(lowu, pre[v]);}}if (fa < 0 && child == 1) iscut[u] = 0;return low[u] = lowu;}void findBCC () {cntlock = cntbcc = 0;memset(pre, 0, sizeof(pre));memset(iscut, 0, sizeof(iscut));memset(bccno, 0, sizeof(bccno));for (int i = 1; i <= N; i++)if (!pre[i]) dfs(i, -1);}void init () {int u, v;memset(vis, 0, sizeof(vis));for (int i = 0; i <= N; i++) G[i].clear();while (M--) {scanf("%d%d", &u, &v);vis[u][v] = vis[v][u] = 1;}for (int i = 1; i <= N; i++) {for (int j = 1; j < i; j++) {if (vis[i][j]) continue;G[i].push_back(j);G[j].push_back(i);}}findBCC();}int solve () {memset(odd, 0, sizeof(odd));for (int i = 1; i <= cntbcc; i++) {for (int j = 0; j < BCC[i].size(); j++)bccno[BCC[i][j]] = i;memset(color, 0, sizeof(color));color[BCC[i][0]] = 1;if (!bipartite(BCC[i][0], i)) {for (int j = 0; j < BCC[i].size(); j++) odd[BCC[i][j]] = 1;}}int ans = 0;for (int i = 1; i <= N; i++)if (!odd[i]) ans++;return ans;}int main () {while (scanf("%d%d", &N, &M) == 2 && N + M) {init ();printf("%d\n", solve());}return 0;}

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