There are nn cities and mm 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 nn, mm and ss (1≤n≤5000,0≤m≤5000,1≤s≤n1≤n≤5000,0≤m≤5000,1≤s≤n) — the number of cities, the number of roads and the index of the capital. Cities are indexed from 11 to nn.
The following mm lines contain roads: road ii is given as a pair of cities uiui, vivi (1≤ui,vi≤n1≤ui,vi≤n, ui≠viui≠vi). For each pair of cities (u,v)(u,v), there can be at most one road from uu to vv. Roads in opposite directions between a pair of cities are allowed (i.e. from uu to vv and from vv to uu).
Output
Print one integer — the minimum number of extra roads needed to make all the cities reachable from city ss. If all the cities are already reachable from ss, print 0.
Examples
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
3
5 4 5
1 2
2 3
3 4
4 1
1
Note
The first example is illustrated by the following:
For example, you can add roads (6,46,4), (7,97,9), (1,71,7) to make all the cities reachable from s=1s=1.
The second example is illustrated by the following:
In this example, you can add any one of the roads (5,15,1), (5,25,2), (5,35,3), (5,45,4) to make all the cities reachable from s=5s=5.
题意：
给定n个节点，M个有向边，和一个节点s。
问最小需要加多少个有向边可以使全部的节点都有到达s节点的路径。
思路：
把除了s节点的其他节点都缩成强连通分量，强连通分量不能到达s节点的，这个分量多加一个边即可到达。
缩成强连通分量的方法可以用dfs+并查集。
枚举每一个边的两边的节点a和b，如果a和b所在的集合（并查集维护出的集合）有一个边联通，那么把这两个节点对应的集合合并（并查集处理合并集合。）
时间复杂度：O（n*m）
细节见代码：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #include <iomanip> #define ALL(x) (x).begin(), (x).end() #define dll(x) scanf("%I64d",&x) #define xll(x) printf("%I64d ",x) #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), ' ', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define db(x) cout<<"== [ "<<x<<" ] =="<<endl; using namespace std; typedef long long ll; ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;} inline void getInt(int* p); const int maxn=10010; const int inf=0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ std::vector<int> v[maxn]; int n,m,rt; int a,b; int par[maxn]; void init() { repd(i,1,n) { par[i]=i; } } int findpar(int x) { if(par[x]==x) { return x; }else { return par[x]=findpar(par[x]); } } int vis[maxn]; bool check(int x,int y) { int res=0; if(x==y) { return 1; } vis[x]=1; for(auto a:v[x]) { if(!vis[a]) { if(check(a,y)) { res=1; break; } } } return res; } int cnt[maxn]; int u[maxn]; int vv[maxn]; int main() { //freopen("D:\common_text\code_stream\in.txt","r",stdin); //freopen("D:\common_text\code_stream\out.txt","w",stdout); gbtb; cin>>n>>m>>rt; repd(i,1,m) { cin>>a>>b; u[i]=a; vv[i]=b; v[a].pb(b); } init(); repd(i,1,m) { MS0(vis); a=u[i]; b=vv[i]; a=findpar(a); b=findpar(b); if(a!=b&&b!=rt&&check(a,b)) { par[a]=b; } } int ans=0; repd(i,1,n) { if(i!=rt) { a=findpar(i); if(a==i) { ans++; } } } cout<<ans<<endl; return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ' ' || ch == ' '); if (ch == '-') { *p = -(getchar() - '0'); while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 - ch + '0'; } } else { *p = ch - '0'; while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 + ch - '0'; } } }