ch6092 hide and seek (DAG minimum coverage points repeatable path)

Ideas:

Why this topic is the smallest positive solution path DAG repeatable point cover it? ? ? Bloggers depressed for two days (actually playing the game for two days) , get the answer suddenly from the game (actually playing the game dishes, abused, or encountered anchor ,,, autistic, and will come to learn child)

Small BB, text from this moment. 

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First clear, solving one of the biggest set of points, did not meet the set path of points between any two points. ohhhh? ? ? This is not the maximum independent set it? ? ? Unfortunately, this is a directed graph, the largest independent set is for free, and if you go to define a bipartite graph, you will find the map, the front has a premise, a undirected graph how how, that is a bipartite graph in terms of undirected graph. That can not have this as the largest independent set to consider a plan.

But we have to know waypoint FIG minimum coverage, defined as follows: select at least disjoint edge covering all vertices . Minimum path covered can be repeated: select at least cover all the edges may intersect the vertices . For this set of paths, each picked out from the inside up to a point, if you pick out more than one point, for example two, then certainly there is a simple path may be a connection between these two points, that is certainly both one party can reach the other side. So from all of our paths, each path pick a point, finally make up the largest collection point.

There is little need to explain why this issue is covered by a minimum path can be repeated, because the subject has said that if a point of walking along a path to reach another point, the two points also could see of each other , that is to say for the side a-> b, b-> c, indirect a-> c can be considered. Therefore, on the first request at floyd transitive closure split point to establish a bipartite graph, DAG and then run again covered a minimum path point. (Such problems are more accustomed to bloggers adjacency matrix directly QAQ)

 

/**
 *
 * Author: correct
 *
 */
#include <bits/stdc++.h>
#define mem(a, b) memset(a, b, sizeof a)
using namespace std;
const int N = 210;
bool w[N][N];
int match[N];
bool vis[N];
int n, m;
bool dfs(int x){
	for (int i = 1; i <= n; i++){
		if (w[x][i] && !vis[i]){
			vis[i] = 1;
			if (match[i] == -1 || dfs(match[i])){
				match[i] = x;
				return 1;
			}
		}
	}
	return 0;
}
void pre_work(){
	mem(match, -1);
	mem(vis, 0);
	for (int k = 1; k <= n; k++){
		for (int i = 1; i <= n; i++){
			for (int j = 1; j <= n; j++){
				w[i][j] = (w[i][k] && w[k][j]) || w[i][j];
			}
		}
	}
}
int solve(){
	int ans = n;
	for (int i = 1; i <= n; i++){
		mem(vis, 0);
		if (dfs(i))ans--;
	}
	return ans;
}
int main()
{
	mem(w, 0);
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= m; i++){
		int x, y;
		scanf("%d %d", &x, &y);
		w[x][y] = 1;
	}
	pre_work();
	printf("%d\n", solve());
	return 0;
}