【POJ3648】Wedding

【题目链接】

• 点击打开链接

【思路要点】

• 需要输出方案的 $2-SAT$ 问题模板题。
• 时间复杂度 $O(N+M)$ 。

【代码】

#include
#include
#include
#include
#include
#include
using namespace std;
const int MAXN = 105;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {
      x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {
      x = min(x, y); } 
template <typename T> void read(T &x) {
      
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	if (c == 'h') x = x * 2 - 1;
	if (c == 'w') x = x * 2 + 0;
	x *= f;
}
template <typename T> void write(T x) {
      
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
      
	write(x);
	puts("");
}
int n, m, p[MAXN];
vector <int> a[MAXN], b[MAXN];
bool instack[MAXN];
int top, Stack[MAXN], tot, belong[MAXN];
int timer, dfn[MAXN], low[MAXN];
void work(int pos) {
      
	Stack[++top] = pos;
	instack[pos] = true;
	dfn[pos] = low[pos] = ++timer;
	for (unsigned i = 0; i < a[pos].size(); i++)
		if (dfn[a[pos][i]] == 0) {
      
			work(a[pos][i]);
			chkmin(low[pos], low[a[pos][i]]);
		} else if (instack[a[pos][i]]) chkmin(low[pos], dfn[a[pos][i]]);
	if (dfn[pos] == low[pos]) {
      
		int tmp = Stack[top--];
		belong[tmp] = ++tot;
		instack[tmp] = false;
		while (tmp != pos) {
      
			tmp = Stack[top--];
			belong[tmp] = tot;
			instack[tmp] = false;
		}
	}
}
bool ban[MAXN];
void dfs(int pos) {
      
	if (ban[pos]) return;
	ban[pos] = true;
	for (unsigned i = 0; i < b[pos].size(); i++)
		dfs(b[pos][i]);
}
void solve() {
      
	timer = tot = 0;
	memset(dfn, 0, sizeof(dfn));
	for (int i = 1; i <= 2 * n; i++)
		if (dfn[i] == 0) work(i);
	for (int i = 1; i <= 2 * n; i++)
		if (belong[i] == belong[p[i]]) {
      
			puts("bad luck");
			return;
		}
	static int d[MAXN], q[MAXN], pr[MAXN];
	for (int i = 1; i <= tot; i++) {
      
		b[i].clear();
		d[i] = 0;
	}
	for (int i = 1; i <= 2 * n; i++)
		pr[belong[i]] = belong[p[i]];
	for (int i = 1; i <= 2 * n; i++)
	for (unsigned j = 0; j < a[i].size(); j++)
		if (belong[a[i][j]] != belong[i]) {
      
			b[belong[a[i][j]]].push_back(belong[i]);
			d[belong[i]]++;
		}
	int l = 1, r = 0;
	for (int i = 1; i <= 2 * n; i++)
		if (d[i] == 0) q[++r] = i;
	while (l <= r) {
      
		int tmp = q[l++];
		for (unsigned i = 0; i < b[tmp].size(); i++)
			if (--d[b[tmp][i]] == 0) {
      
				q[++r] = b[tmp][i];
			}
	}
	static bool sel[MAXN];
	memset(sel, false, sizeof(sel));
	memset(ban, false, sizeof(ban));
	for (int i = 1; i <= r; i++)
		if (!ban[i]) {
      
			sel[i] = true;
			dfs(pr[i]);
		}
	for (int i = 1; i <= n; i++) {
      
		assert(sel[belong[i * 2 - 1]] ^ sel[belong[i * 2 + 0]]);
		if (sel[belong[i * 2 - 1]]) printf("%dh ", i);
		else printf("%dw ", i);
	}
	printf("\n");
}
int main() {
      
	read(n), read(m);
	while (n != 0 || m != 0) {
      
		n--;
		for (int i = 1; i <= 2 * n; i++)
			a[i].clear();
		for (int i = 1; i <= n; i++) {
      
			p[i * 2 - 1] = i * 2 + 0;
			p[i * 2 + 0] = i * 2 - 1;
		}
		for (int i = 1; i <= m; i++) {
      
			int x, y; read(x), read(y);
			if (x == 0 || y == 0) continue;
			if (x == -1) {
      
				a[p[y]].push_back(y);
				continue;
			}
			if (y == -1) {
      
				a[p[x]].push_back(x);
				continue;
			}
			a[p[y]].push_back(x);
			a[p[x]].push_back(y);
		}
		solve();
		read(n), read(m);
	}
	return 0;
}