2017ICPC南宁M The Maximum Unreachable Node Set （偏序集最长反链）

题意：给你一张DAG，让你选取最多的点，使得这些点之间互相不可达。

思路：此问题和最小路径可重复点覆盖等价，先在原图上跑一边传递闭包，然后把每个点拆成两个点i, i + n, 原图中的边(a, b)变成(a, b + n)，跑一变网络流, 答案就是n - maxflow;

代码：

#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 305;
const int maxm = 100010;
bitset<maxn> b[maxn];
queue<int> q;
int head[maxn], ver[maxm], Next[maxm], edge[maxm], d[maxn];
vector<int> G[maxn];
int n, m, s, t, tot, maxflow;
bool v[maxn];
void dfs(int x) {
	if(v[x]) return;
	b[x][x] = 1;
	for (auto y : G[x]) {
		dfs(y);
		b[x] |= b[y];
	}
    v[x] = 1;
	return;
}
void add(int x, int y, int z) {
    ver[++tot] = y, edge[tot] = z, Next[tot] = head[x], head[x] = tot;
    ver[++tot] = x, edge[tot] = 0, Next[tot] = head[y], head[y] = tot;
}
bool bfs() {
	memset(d, 0, sizeof(d));
	while(q.size()) q.pop();
	q.push(s);d[s] = 1;
	while(q.size()) {
		int x=  q.front();
		q.pop();
		for (int i = head[x]; i; i = Next[i]) {
			if(edge[i] && !d[ver[i]]) {
				q.push(ver[i]);
				d[ver[i]] = d[x] + 1;
				if(ver[i] == t) return 1;
			}
		}
    }
	return 0;
}

int dinic(int x, int flow) {
	if(x == t) return flow;
	int rest = flow, k;
	for (int i = head[x]; i && rest; i = Next[i]) {
		if(edge[i] && d[ver[i]] == d[x] + 1) {
			k = dinic(ver[i], min(rest, edge[i]));
			if(!k) d[ver[i]] = 0;
			edge[i] -= k;
			edge[i ^ 1] += k;
			rest -= k;
		}
	}
	return flow - rest;
}
int main() {
	int T, x, y;
	scanf("%d", &T);
	while(T--) {
		scanf("%d%d", &n, &m);
        tot = 1;
        s = n * 2 + 1, t = n * 2 + 2;
        for (int i = 1; i <= n; i++)
            b[i].reset();
        memset(head, 0, sizeof(head));
		for (int i = 1; i <= n; i++) {
			G[i].clear();
			v[i] = 0;
		}
		maxflow = 0;
		for (int i = 1; i <= m; i++) {
			scanf("%d%d", &x, &y);
			G[x].push_back(y);
		}
		for (int i = 1; i <= n; i++) {
			if(!v[i]) {
				dfs(i);
			}
		}
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				if(i == j) continue;
				if(b[i][j] == 1) {
					add(i, j + n, 1);
				}
			}
		}
		for (int i = 1; i <= n; i++) {
			add(s, i, 1);
			add(i + n, t, 1);
		}
		int flow = 0;
		while(bfs())
			while(flow = dinic(s, INF)) maxflow += flow;
		printf("%d\n", n - maxflow);
	}
	return 0;
}