Light OJ 1429 Assassin`s Creed (II) BFS+缩点+最小路径覆盖

题目来源：Light OJ 1429 Assassin`s Creed (II)

题意：最少几个人走全然图 能够反复走 有向图

思路：假设是DAG图而且每一个点不能反复走 那么就是裸的最小路径覆盖 如今不是DAG 可能有环 而且每一个点可能反复走 对于有环 能够缩点 缩点之后的图是DAG图 另外点能够反复走和POJ 2594一样 先预处理连通性

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
const int maxn = 1010;


int vis[maxn];
int y[maxn];
vector <int> G[maxn], G2[maxn], G3[maxn];
int n, m;
int a[maxn][maxn];

int pre[maxn];
int low[maxn];
int sccno[maxn];
int dfs_clock;
int scc_cnt;
stack <int> S;

void dfs(int u)
{
	pre[u] = low[u] = ++dfs_clock;
	S.push(u);
	for(int i = 0; i < G[u].size(); i++)
	{
		int v = G[u][i];
		if(!pre[v])
		{
			dfs(v);
			low[u] = min(low[u], low[v]);
		}
		else if(!sccno[v])
			low[u] = min(low[u], pre[v]);
	}
	if(pre[u] == low[u])
	{
		scc_cnt++;
		while(1)
		{
			int x = S.top();
			S.pop();
			sccno[x] = scc_cnt;
			if(x == u)
				break;
		}
	}
}
void find_scc()
{
	dfs_clock = scc_cnt = 0;
	memset(sccno, 0, sizeof(sccno));
	memset(pre, 0, sizeof(pre));
	for(int i = 1; i <= n; i++)
		if(!pre[i])
			dfs(i);
}

void BFS(int u)
{
	queue <int> Q;
	memset(vis, 0, sizeof(vis));	
	vis[u] = true;
	Q.push(u);
	while(!Q.empty())
	{
		int x = Q.front(); Q.pop();
		for(int i = 0; i < G2[x].size(); i++)
		{
			int v = G2[x][i];
			if(vis[v])
				continue;
			vis[v] = true;
			G[u].push_back(v);
			Q.push(v);
		}
	}
}
bool dfs2(int u)
{
    for(int i = 0; i < G3[u].size(); i++)
    {
        int v = G3[u][i];
        if(vis[v])
            continue;
        vis[v] = true;
        if(y[v] == -1 || dfs2(y[v]))
        {
            y[v] = u;
            return true;
        }
    }
    return false;
}
int match()
{
    int ans = 0;
    memset(y, -1, sizeof(y));
    for(int i = 1; i <= scc_cnt; i++)
    {
		memset(vis, 0, sizeof(vis));
		if(dfs2(i))
		    ans++;
    }
    return ans;
}

int main()
{
	int cas = 1;
	int T;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d %d", &n, &m);
		for(int i = 0; i <= n; i++)
			G[i].clear(), G2[i].clear(), G3[i].clear();
		while(m--)
		{
			int u, v;
			scanf("%d %d", &u, &v);
			G2[u].push_back(v);
		}
		for(int i = 1; i <= n; i++)
			BFS(i);
		find_scc();
		
		for(int u = 1; u <= n; u++)
		{
			for(int i = 0; i < G[u].size(); i++)
			{
				int v = G[u][i];
				if(sccno[u] != sccno[v])
					G3[sccno[u]].push_back(sccno[v]);
			}
		}
		printf("Case %d: %d\n", cas++, scc_cnt-match());
	}
    return 0;
}