JZOJ 3423.Vani和Cl2捉迷藏 & [CTSC2008]祭祀

\(\text{Problem}\)

求一个 \(DAG\) 的最长反链

\(\text{Solution}\)

由 \(Dilworth\) 定理只最长反链等于最小链覆盖

而原图的链是可相交的，所以我们先做一遍 \(Floyd\) 传递闭包，使得原图的链不必相交即可覆盖

这样就转化为最小链覆盖（顶点不可相交）

于是用网络流经典模型解决

\(\text{Code}\)

#include<cstdio>
#include<iostream>
using namespace std;

const int N = 405;
int n, m, h[N], S, T, g[N][N];

struct edge{
	int to, nxt, w;
}e[100005];
inline void add(int u, int v, int w)
{
	static int tot = 1;
	e[++tot] = edge{v, h[u], w}, h[u] = tot;
}

int Q[N], cur[N], dep[N];
inline int bfs()
{
	for(int i = S; i <= T; i++) cur[i] = h[i], dep[i] = 0;
	int head = 0, tail = 1;
	Q[1] = S, dep[S] = 1;
	while (head < tail)
	{
		int now = Q[++head];
		for(int i = h[now]; i; i = e[i].nxt)
		{
			int v = e[i].to;
			if (dep[v] || !e[i].w) continue;
			dep[v] = dep[now] + 1, Q[++tail] = v;
		}
	}
	return dep[T];
}
int dfs(int x, int mi)
{
	if (x == T || mi <= 0) return mi;
	int flow = 0;
	for(int i = cur[x]; i; i = e[i].nxt)
	{
		cur[x] = i;
		int v = e[i].to;
		if (dep[v] ^ (dep[x] + 1) || !e[i].w) continue;
		int f = dfs(v, min(mi, e[i].w));
		if (f <= 0) continue;
		flow += f, mi -= f, e[i].w -= f, e[i ^ 1].w += f;
		if (mi <= 0) break;
	}
	return flow;
}

int dinic()
{
	int res = 0;
	while (bfs()) res += dfs(S, N);
	return res;
}

int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 1, x, y; i <= m; i++) scanf("%d%d", &x, &y), g[x][y] = 1;
	for(int k = 1; k <= n; k++)
		for(int i = 1; i <= n; i++)
		if (g[i][k])
			for(int j = 1; j <= n; j++)
			if (g[k][j]) g[i][j] = 1;
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
			if (g[i][j]) add(i, j + n, 1), add(j + n, i, 0);
	T = 2 * n + 1;
	for(int i = 1; i <= n; i++) add(S, i, 1), add(i, S, 0), add(i + n, T, 1), add(T, i + n, 0);
	printf("%d\n", n - dinic());
}