Reachability from the Capital

问题描述：

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.

问题分析：本道题的大意是给你一张有向图，需要使得给定的点可以到达图中任意的一个节点位置，问至少有在图中修多少条路。

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 $s=1$.

本道题的解题思想主要是dfs。首先我们遍历1到n个点，把到达过的位置做一个标记，并将每个点可以到达的位置保存到堆栈中。然后标记清零，bfs定点s可以到达的节点，做标记。堆栈中未被标记的位置便是需要建立通路的位置。

AC代码：

#include <iostream>
#include<iomanip>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<set>
#include<stack>
#include<queue>
using namespace std;

const int N = 5005;

int m,n,s;
int a[N];
int used[N];
vector<int>G[N];
stack<int>st;
//遍历每个点所能达到的节点位置，并保存 
void dfs(int u){
	used[u] = 1;
	/*
	for(int v:G[u])
        if(!vis[v])
            dfs(v);
    st.push(u);
    */
	for(int i = 0;i < G[u].size();i++){
		int v = G[u][i];
		if(!used[v]){
			dfs(v);
		}
	}
	st.push(u);
}

void dfs1(int v){
	used[v] = 1;
	for(int i = 0;i < G[v].size();i++){
		int u = G[v][i];
		if(!used[u]){
			dfs(u);
		}
	}
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n>>m>>s;
	int a,b;
	memset(used,0,sizeof(used));
	for(int i = 1;i <= m;i++){
		cin>>a>>b;
		//从a到b有路 
		G[a].push_back(b);
	}
	for(int i= 1;i <= n;i++){
		if(!used[i]){
			//遍历每个点 
			dfs(i);
		}
	}
	memset(used,0,sizeof(used));
	dfs1(s);
	int cnt = 0;
	while(!st.empty()){
		int u = st.top();
		st.pop();
		if(!used[u]){
			dfs1(u);
			cnt++;
		} 
	}
	cout<<cnt<<endl;
	return 0;
}