UVA 1364 - Knights of the Round Table

题目链接

题意：有n个圆桌骑士，知道一些骑士互相憎恨，现在要开圆桌会议，每次最少3个人，必须是奇数人数，并且互相憎恨的骑士不能在相邻，问有多少骑士是一次都无法参加的

思路：把每个骑士可以相邻的连边，然后做双连通分量，然后对于每个连通分量，利用二分图染色判定去判断是否是奇圈

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
using namespace std;

const int N = 1005;

struct Edge {
	int u, v;
	Edge() {}
	Edge(int u, int v) {
		this->u = u;
		this->v = v;
	}
};

int pre[N], bccno[N], dfs_clock, bcc_cnt;
bool iscut[N];

vector<int> g[N], bcc[N];
stack<Edge> S;

int dfs_bcc(int u, int fa) {
	int lowu = pre[u] = ++dfs_clock;
	int child = 0;
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		Edge e = Edge(u, v);
		if (!pre[v]) {
			S.push(e);
			child++;
			int lowv = dfs_bcc(v, u);
			lowu = min(lowu, lowv);
			if (lowv >= pre[u]) {
				iscut[u] = true;
				bcc_cnt++; bcc[bcc_cnt].clear(); //start from 1
				while(1) {
					Edge x = S.top(); S.pop();
					if (bccno[x.u] != bcc_cnt) {bcc[bcc_cnt].push_back(x.u); bccno[x.u] = bcc_cnt;}
					if (bccno[x.v] != bcc_cnt) {bcc[bcc_cnt].push_back(x.v); bccno[x.v] = bcc_cnt;}
					if (x.u == u && x.v == v) break;
				}
			}
		} else if (pre[v] < pre[u] && v != fa) {
			S.push(e);
			lowu = min(lowu, pre[v]);
		}
	}
	if (fa < 0 && child == 1) iscut[u] = false;
	return lowu;
}

void find_bcc(int n) {
	memset(pre, 0, sizeof(pre));
	memset(iscut, 0, sizeof(iscut));
	memset(bccno, 0, sizeof(bccno));
	dfs_clock = bcc_cnt = 0;
	for (int i = 0; i < n; i++)
		if (!pre[i]) dfs_bcc(i, -1);
}

int odd[N], color[N];

bool bipartite(int u, int b) {
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i]; if (bccno[v] != b) continue;
		if (color[v] == color[u]) return false;
		if (!color[v]) {
			color[v] = 3 - color[u];
			if (!bipartite(v, b)) return false;
		}
	}
	return true;
}

int n, m, A[N][N];

int main() {
	int cas = 0;
	while (~scanf("%d%d", &n, &m) && n) {
		for (int i = 0; i < n; i++) g[i].clear();
		memset(A, 0, sizeof(A));
		for (int i = 0; i < m; i++) {
			int u, v;
			scanf("%d%d", &u, &v); u--; v--;
			A[u][v] = A[v][u] = 1;
		}
		for (int u = 0; u < n; u++) {
			for (int v = u + 1; v < n; v++)
				if (!A[u][v]) {
					g[u].push_back(v);
					g[v].push_back(u);
				}
		}
		find_bcc(n);
		memset(odd, 0, sizeof(odd));
		for (int i = 1; i <= bcc_cnt; i++) {
			memset(color, 0, sizeof(color));
			for (int j = 0; j < bcc[i].size(); j++) bccno[bcc[i][j]] = i;
			int u = bcc[i][0];
			color[u] = 1;
			if (!bipartite(u, i)) {
				for (int j = 0; j < bcc[i].size(); j++)
					odd[bcc[i][j]] = 1;
			}
		}
		int ans = n;
		for (int i = 0; i < n; i++)
			ans -= odd[i];
		printf("%d\n", ans);
	}
	return 0;
}

UVA 1364 - Knights of the Round Table（双连通+二分图判定）