BZOJ5335 [TJOI2018]智力竞赛

题目蓝链

Solution

我们可以贪心的考虑，就是尽量把小的覆盖完。我们把所有的点从小到大排序，然后直接二分覆盖前\(k\)个点，用网络流跑一下二分图求出最小链覆盖，然后就判断一下点数减去最小链有没有超过给定的人数

由于这道题的链可以重叠的，所以我们可以直接求出每一个点可达的所有点。然后拆点，对于每一个点和其可达的点，从左边的点连向右边的点。然后需要的最小链的数量就是总点数减去总流量

Code

#include <bits/stdc++.h>

using namespace std;

#define squ(x) ((LL)(x) * (x))
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long LL;
typedef pair<int, int> pii;

inline int read() {
	int sum = 0, fg = 1; char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') fg = -1;
	for (; isdigit(c); c = getchar()) sum = (sum << 3) + (sum << 1) + (c ^ 0x30);
	return fg * sum;
}

const int maxn = 500 + 10;
const int maxm = 5e5 + 10;
const int inf = 0x3f3f3f3f;

int n, m;

namespace Dinic {

	int S, T;
	int Begin[maxn << 1], Next[maxm + (maxn << 2)], To[maxm + (maxn << 2)], w[maxm + (maxn << 2)], e;
	int d[maxn << 1], cur[maxn << 1];

	void init() {
		S = 0, T = m << 1 | 1, e = -1;
		memset(Begin, -1, sizeof Begin);
	}

	void add(int x, int y) {
		To[++e] = y, Next[e] = Begin[x], Begin[x] = e, w[e] = 1;
		To[++e] = x, Next[e] = Begin[y], Begin[y] = e, w[e] = 0;
	}

	int bfs() {
		memset(d, 0, sizeof d); d[S] = 1;
		queue<int> q; q.push(S);
		while (!q.empty()) {
			int now = q.front(); q.pop();
			for (int i = Begin[now]; i + 1; i = Next[i]) {
				int son = To[i];
				if (!d[son] && w[i] > 0) {
					d[son] = d[now] + 1; q.push(son);
				}
			}
		}
		return d[T];
	}

	int dfs(int now, int flow) {
		if (now == T) return flow;
		int sum = 0;
		for (int &i = cur[now]; i + 1; i = Next[i]) {
			int son = To[i];
			if (w[i] > 0 && d[son] == d[now] + 1) {
				int l = dfs(son, min(flow, w[i]));
				if (l) {
					w[i] -= l, w[i ^ 1] += l;
					flow -= l, sum += l;
					if (!flow) break;
				}
			}
		}
		return sum;
	}

	int solve() {
		for (int i = 1; i <= m; i++) add(S, i), add(i + m, T);
		int ans = 0;
		while (bfs()) {
			for (int i = S; i <= T; i++) cur[i] = Begin[i];
			ans += dfs(S, inf);
		}
		return ans;
	}
}

struct node {
	int x, id;
	bool operator < (const node &t) const { return x < t.x; }
}A[maxn];

bool b[maxn][maxn];

bool check(int pos) {
	Dinic::init();
	for (int i = 1; i <= pos; i++)
		for (int j = 1; j <= pos; j++) {
			int x = A[i].id, y = A[j].id;
			if (i == j || !b[x][y]) continue;
			Dinic::add(x, y + m);
		}
	int res = Dinic::solve();
	return (pos - res) <= n;
}

int main() {
#ifdef xunzhen
	freopen("comp.in", "r", stdin);
	freopen("comp.out", "w", stdout);
#endif

	n = read() + 1, m = read();
	for (int i = 1; i <= m; i++) {
		b[i][i] = 1; A[i] = (node){read(), i};
		int k = read();
		for (int j = 1; j <= k; j++) b[i][read()] = 1;
	}

	for (int k = 1; k <= m; k++)
		for (int i = 1; i <= m; i++)
			for (int j = 1; j <= m; j++)
				b[i][j] |= b[i][k] & b[k][j];

	sort(A + 1, A + m + 1);
	int l = 1, r = m;
	while (l <= r) {
		int mid = (l + r) >> 1;
		if (check(mid)) l = mid + 1;
		else r = mid - 1;
	}

	if (l <= m) printf("%d\n", A[l].x);
	else puts("AK");

	return 0;
}

Summary

第一次做最小链覆盖的题，感觉不是太难，过了样例就直接A了