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 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 vvto 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
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 (6,46,4), (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.
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
using namespace std;
#define LL long long
const int MAXN = 1e6+10;
const int INF = 0x3f3f3f3f;
const int N = 10010;
int n,m,s;
int tot,num,temp;
int dfn[N],low[N],color[N],vis[N],in[N];
vector<int>G[N];
stack<int>S;
void init(){
tot = num = temp = 0;
while(!S.empty()) S.pop();
for(int i=0;i<=n;i++){
G[i].clear();
}
memset(dfn,0,sizeof(dfn));
memset(vis,0,sizeof(vis));
memset(low,0,sizeof(low));
memset(in,0,sizeof(in));
memset(color,0,sizeof(color));
}
void tarjan(int x){
low[x] = dfn[x] = ++ tot;
S.push(x);
vis[x] = 1;
for(int i=0;i<G[x].size();i++){
int v = G[x][i];
if(!dfn[v]){
tarjan(v);
low[x] = min(low[x],low[v]);
}
else if(vis[v])
low[x] = min(low[x],dfn[v]);
}
if(dfn[x] == low[x]){
num++;
while(1){
int now = S.top();
S.pop();
color[now] = num;
vis[now] = 0;
if(now == x) break;
}
}
}
void Build(){
for(int i=1;i<=n;i++){
for(int j=0;j<G[i].size();j++){
int v = G[i][j];
if(color[v] != color[i]){
in[color[v]] ++;
}
}
}
for(int i=1;i<=num;i++){
if(in[i]==0){
temp ++;
}
}
}
int main(){
while(scanf("%d%d%d",&n,&m,&s)!=EOF){
init();
int u,v;
for(int i=0;i<m;i++){
scanf("%d%d",&u,&v);
G[u].push_back(v);
}
for(int i=1;i<=n;i++){
if(!dfn[i]) tarjan(i);
}
Build();
if(in[color[s]]==0) temp--;
cout<<temp<<endl;
}
return 0;
}