Reachability from the Capital

题目描述

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

Input

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

Output

3

Input

5 4 5
1 2
2 3
3 4
4 1

Output

1

 

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.

 

题解：

强连通缩点后统计入度为0的个数ans，然后看首都的入度是否为0；如果是则ans-1;

 #include<cstdio>
 #include <algorithm>
 #include <stack>
 #include <vector>
 #include <cstring>
 using namespace std;

 const int MAXN=1e5+;
 const int inf=0x3f3f3f3f;
 struct node{
     int to;
     int next;
 }edge[MAXN*];
 int head[MAXN];
 int val[MAXN];
 bool instack[MAXN];
 int cnt;
 int dfn[MAXN],low[MAXN];
 int sum[MAXN];
 void add(int x,int y)
 {
     edge[++cnt].to =y;
     edge[cnt].next=head[x];
     head[x]=cnt;
 }
 int Time,num;
 stack<int >st;
 int du[MAXN];
 int color[MAXN];
 int x[MAXN],y[MAXN];
 void tarjan(int u)
 {
     dfn[u]=low[u]= ++Time;
     st.push(u);
     instack[u]=true;
     for (int i = head[u]; i !=- ; i=edge[i].next) {
         int v=edge[i].to;
         if(!dfn[v]){
             tarjan(v);
             low[u]=min(low[u],low[v]);
         }
         else if(instack[v]) low[u]=min(low[u],dfn[v]);
     }
     if(dfn[u]==low[u])
     {
         int x;
         num++;
         while() {
             x=st.top();
             st.pop();
             color[x]=num;
             instack[x]=false;
             if(x==u) break;
         }

     }
 }

 int main()
 {
     int n,m,s;
     scanf("%d%d%d",&n,&m,&s);
     cnt=;
     memset(head,-,sizeof(head));
     memset(instack,false, sizeof(instack));
     memset(sum, ,sizeof(sum));
     for (int i = ; i <=m ; ++i) {
         scanf("%d%d",&x[i],&y[i]);
         add(x[i],y[i]);
     }
     for (int i = ; i <=n ; ++i) {
         if(!dfn[i]) tarjan(i);
     }
     for (int i = ; i <=m ; ++i) {
         if(color[x[i]]!=color[y[i]])
         {
             du[color[y[i]]]++;
         }
     }

     int ans=;
     for (int i = ; i <=num ; ++i) {
         if(du[i]==) ans++;
     }
     if(du[color[s]]==) ans--;
     printf("%d\n",ans);
     return ;
 }