Codeforces Round 970 (Div. 3) D. Sakurako‘s Hobby

链接

cf_Sakurako‘s Hobby

大意：

给一堆点和边，并给出点的颜色，输出每个点能遍历到几个黑点

思路：

1、这些点边里面有拓扑结构， 也有环

2、先处理拓扑排序的一些点，依次遍历无父节点的即可，之后就会剩下环

3、有环的说明每个点都能去到环内任意一点，那么直接就记录一个sum，然后递归求环内黑点，最后把sum赋值给环内所有点即可

代码

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 10;


int e[N], ne[N], h[N], idx;

void add(int a, int b)
{
	e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

int p[N];
int vis[N];
int in[N];
int f[N];
vector<int> circle; // 为了之后给环里面的东西幅值
int sum;
string s;
void dfs(int u)
{
	circle.push_back(u);
	if (s[u] == '0')sum++;
	vis[u] = 1;
	for (int i = h[u]; i != -1; i = ne[i])
	{
		int j = e[i];
		if (!vis[j])dfs(j);
	}
}

void solve()
{
	memset(h, -1, sizeof h);
	idx = 0;
	int n;cin >> n;
	for (int i = 1; i <= n; i++)in[i] = f[i] = vis[i] = 0;
	for (int i = 1; i <= n; i++)
	{
		cin >> p[i];
		if (i != p[i])
		{
			add(i, p[i]);
			in[p[i]]++;
		}
	}
	cin >> s;
	s = " " + s;
// 处理拓扑排序
	queue<int>q;
	for (int i = 1; i <= n; i++) //拓扑的头 对无环的点
		if (!in[i])q.push(i), vis[i] = 1;

	while (q.size())
	{
		int t = q.front();
		q.pop();
		if (s[t] == '0')f[t] ++;
		for (int i = h[t]; i != -1; i = ne[i])
		{
			int j = e[i];
			in[j] --;
			if (!in[j] && !vis[j])q.push(j), vis[j] = 1;
		}
	}

	for (int i = 1; i <= n; i ++)
		if (!vis[i])
		{
			circle.clear();
			sum = 0; // 记录环中黑点个数
			dfs(i);
			for (auto it : circle)f[it] = sum;
		}
	for (int i = 1; i <= n; i++)
		cout << f[i] << ' ';
	cout << endl;
}

signed main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int t = 1;
	cin >> t;
	
	while (t--) solve();
	return 0;
}