【BZOJ5335】【TJOI2018】智力竞赛

【题目链接】

• 点击打开链接
  

【思路要点】

• 二分答案，然后用有上下界的最小流验证答案。
• 时间复杂度\(O(LogM*Dinic(M,\sum K_i))\)。

【代码】

#include
using namespace std;
const int MAXN = 505;
const int MAXP = 2005;
const int INF = 1e9;
template  void chkmax(T &x, T y) {x = max(x, y); }
template  void chkmin(T &x, T y) {x = min(x, y); } 
template  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';
	x *= f;
}
template  void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template  void writeln(T x) {
	write(x);
	puts("");
}
struct edge {int dest; int flow; unsigned home; };
int s, t, olds, oldt, tot, dist[MAXP];
vector  a[MAXP];
unsigned curr[MAXP];
int dinic(int pos, int limit) {
	if (pos == t) return limit;
	int used = 0, tmp;
	for (unsigned &i = curr[pos]; i < a[pos].size(); i++)
		if (a[pos][i].flow != 0 && dist[pos] + 1 == dist[a[pos][i].dest] && (tmp = dinic(a[pos][i].dest, min(limit - used, a[pos][i].flow))) != 0) {
			used += tmp;
			a[pos][i].flow -= tmp;
			a[a[pos][i].dest][a[pos][i].home].flow += tmp;
			if (used == limit) return used;
		}
	return used;
}
bool bfs() {
	static int q[MAXP], l = 0, r = -1;
	for (int i = 0; i <= r; i++)
		dist[q[i]] = 0;
	q[l = r = 0] = s, dist[s] = 1;
	while (l <= r) {
		int tmp = q[l++];
		for (unsigned i = 0; i < a[tmp].size(); i++)
			if (a[tmp][i].flow != 0 && dist[a[tmp][i].dest] == 0) {
				q[++r] = a[tmp][i].dest;
				dist[a[tmp][i].dest] = dist[tmp] + 1;
			}
	}
	return dist[t] != 0;
}
void addedge(int s, int t, int flow) {
	a[s].push_back((edge) {t, flow, a[t].size()});
	a[t].push_back((edge) {s, 0, a[s].size() - 1});
}
vector  e[MAXN];
int limit, n, val[MAXN], pos[MAXN];
int inp[MAXN], outp[MAXN];
bool cmp(int x, int y) {
	return val[x] < val[y];
}
bool check(int mid) {
	tot = 0;
	olds = ++tot; a[tot].clear();
	oldt = ++tot; a[tot].clear();
	s = ++tot; a[tot].clear();
	t = ++tot; a[tot].clear();
	for (int i = 1; i <= n; i++) {
		int tmp = pos[i];
		inp[tmp] = ++tot; a[tot].clear();
		outp[tmp] = ++tot; a[tot].clear();
		addedge(olds, inp[tmp], INF);
		addedge(outp[tmp], oldt, INF);
		addedge(inp[tmp], outp[tmp], INF);
		if (i <= mid) {
			addedge(s, outp[tmp], 1);
			addedge(inp[tmp], t, 1);
		}
	}
	for (int i = 1; i <= n; i++) {
		int tmp = pos[i];
		for (unsigned j = 0; j < e[tmp].size(); j++)
			addedge(outp[tmp], inp[e[tmp][j]], INF);
	}
	int tmp = 0;
	while (bfs()) {
		memset(curr, 0, sizeof(curr));
		tmp += dinic(s, INF);
	}
	addedge(oldt, olds, INF);
	int ans = 0;
	while (bfs()) {
		memset(curr, 0, sizeof(curr));
		ans += dinic(s, INF);
	}
	return ans <= limit + 1;
}
int main() {
	read(limit), read(n);
	for (int i = 1; i <= n; i++) {
		read(val[i]);
		pos[i] = i;
		int k; read(k);
		while (k--) {
			int x; read(x);
			e[i].push_back(x);
		}
	}
	sort(pos + 1, pos + n + 1, cmp);
	int l = 1, r = n;
	while (l < r) {
		int mid = (l + r + 1) / 2;
		if (check(mid)) l = mid;
		else r = mid - 1;
	}
	if (l == n) printf("AK\n");
	else printf("%d\n", val[l + 1]);
	return 0;
}