999E - Reachability from the Capital(Codeforces Round #490 (Div. 3))

                                                                                here is the problem

E. Reachability from the Capital

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are $n$ cities and $m$ roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.

What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?

New roads will also be one-way.

Input

The first line of input consists of three integers $n$, $m$ and $s$ ($1\le n\le 5000,0\le m\le 5000,1\le s\le n$) — the number of cities, the number of roads and the index of the capital. Cities are indexed from $1$ to $n$.

The following $m$ lines contain roads: road $i$ is given as a pair of cities ,  ($1\le {}^{},{}^{}\le n$, ${}^{}\ne {}^{}$). For each pair of cities $(u,v)$, there can be at most one road from $u$ to $v$. Roads in opposite directions between a pair of cities are allowed (i.e. from $u$ to $v$ and from $v$ to $u$).

Output

Print one integer — the minimum number of extra roads needed to make all the cities reachable from city $s$. If all the cities are already reachable from $s$, print 0.

Examples

input

Copy

9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1

output

Copy

3

input

Copy

5 4 5
1 2
2 3
3 4
4 1

output

Copy

1

Note

The first example is illustrated by the following:

For example, you can add roads ($6,4$), ($7,9$), ($1,7$) to make all the cities reachable from $s=1$.

The second example is illustrated by the following:

In this example, you can add any one of the roads ($5,1$), ($5,2$), ($5,3$), ($5,4$) to make all the cities reachable from $s=5$.

solution:

This problem is (almost) equivalent to the following: count the number of sources (the vertices with indegree equal to $0$) in the given graph's condensation. Thus, there exist solutions with complexity $O(n+m)$. However, the constraints in the problem are small, so solutions with complexity $O(n\cdot m)$ also pass.

One of these solutions is the following: first, let's mark all the vertices reachable from $s$ as good, using a simple DFS. Then, for each bad vertex $v$, count the number of bad vertices reachable from $v$ (it also can be done by simple DFS). Let this number be . Now, iterate over all bad vertices in non-increasing order of . For the current bad vertex $v$, if it is still not marked as good, run a DFS from it, marking all the reachable vertices as good, and increase the answer by $1$ (in fact, we are implicitly adding the edge $(s,v)$). It can be proved that this solution gives an optimal answer.

（copy from here）

算了，不秀英语了，下面进入正题。

这一题的思路其实也不难（但比赛写了个缩点被卡shi）

这题就是dfs，只不过要用敏锐目光察觉出来

上代码：  其实也不难理解，看看代码就懂了

#include <bits/stdc++.h>

using namespace std;

const int N = 5010;

int n, m, s;
vector<int> g[N];
bool used[N];
bool ok[N];

int cnt;

void dfs1(int v) {
	ok[v] = true;
	for (auto to : g[v])
		if (!ok[to])
			dfs1(to);
}

void dfs2(int v) {
	used[v] = true;
	++cnt;
	for (auto to : g[v])
		if (!used[to] && !ok[to])
			dfs2(to);
}

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	cin >> n >> m >> s;
	--s;
	for (int i = 0; i < m; ++i) {
		int x, y;
		cin >> x >> y;
		--x, --y;
		g[x].push_back(y);
	}
	
	dfs1(s);
	
	vector<pair<int, int>> val;
	for (int i = 0; i < n; ++i) {
		if (!ok[i]) {
			cnt = 0;
			memset(used, false, sizeof(used));
			dfs2(i);
			val.push_back(make_pair(cnt, i));
		}
	}
	sort(val.begin(), val.end());
	reverse(val.begin(), val.end());
	
	int ans = 0;
	for (auto it : val) {
		if (!ok[it.second]) {
			++ans;
			dfs1(it.second);
		}
	}
	
	cout << ans << endl;
	
	return 0;
}

详情请进入codeforces