Codeforces Round #286 (Div. 1) B. Mr. Kitayuta's Technology 连通分量 拓扑排序

B. Mr. Kitayuta's Technology

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Shuseki Kingdom is the world's leading nation for innovation and technology. There aren cities in the kingdom, numbered from1 ton.

Thanks to Mr. Kitayuta's research, it has finally become possible to construct teleportation pipes between two cities. A teleportation pipe will connect two cities unidirectionally, that is, a teleportation pipe from cityx
to city y cannot be used to travel from cityy to cityx. The transportation within each city is extremely developed, therefore if a pipe from
cityx to cityy and a pipe from cityy to cityz are both constructed, people will be able to travel from
cityx to cityz instantly.

Mr. Kitayuta is also involved in national politics. He considers that the transportation between them pairs of city(ai, bi)
(1 ≤ i ≤ m) is important. He is planning to construct teleportation pipes so that for each important pair(ai, bi),
it will be possible to travel from cityai to citybi by using one or more teleportation pipes (but
not necessarily from citybi to cityai). Find the minimum number of teleportation pipes that need to
be constructed. So far, no teleportation pipe has been constructed, and there is no other effective transportation between cities.

Input
The first line contains two space-separated integers n andm (2 ≤ n ≤ 105, 1 ≤ m ≤ 105),
denoting the number of the cities in Shuseki Kingdom and the number of the important pairs, respectively.

The following m lines describe the important pairs. Thei-th of them (1 ≤ i ≤ m) contains two space-separated integersai
andbi (1 ≤ ai, bi ≤ n, ai ≠ bi),
denoting that it must be possible to travel from city ai to citybi by using one or more teleportation
pipes (but not necessarily from citybi to cityai). It is guaranteed that all pairs(ai, bi)
are distinct.

Output
Print the minimum required number of teleportation pipes to fulfill Mr. Kitayuta's purpose.

Sample test(s)

Input

4 5
1 2
1 3
1 4
2 3
2 4

Output

3

Input

4 6
1 2
1 4
2 32 4
3 2
3 4

Output

4

ai,bi表示ai能到达bi，要求满足这些条件的图最少用多少条边。

所有有关联的点为一个连通分量，先用DFS划分出所有的连通分量。对于包含n个点的单个连通分量来说，如果它是一个有向无环图，那么用拓扑排序的顺序连起来边数为n-1，能满足条件，并且也是边数最少的，因为n个有关系的点至少也要n-1条边。如果有环的话，那么用n条边把这n个点连成一个环，两两可达。判断是不是有向无环图就用类似拓扑排序的方法。

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<algorithm>
typedef long long LL;
using namespace std;

const int MAXN=100010;

int N,M,idx,belong[MAXN],deg[MAXN],cnt[MAXN];
vector<int> V[MAXN],G[MAXN],Component[MAXN];

void DFS(int u){
belong[u]=idx;
int len=V[u].size();
for(int i=0;i<len;i++){
int v=V[u][i];
if(belong[v]==-1) DFS(v);
}
}

bool DAG(int c){
queue<int> q;
int len=Component[c].size();
for(int i=0;i<len;i++){
int u=Component[c][i];
if(!deg[u]) q.push(u);
}
while(!q.empty()){
int u=q.front();
q.pop();
int len2=G[u].size();
for(int i=0;i<len2;i++){
int v=G[u][i];
deg[v]--;
if(!deg[v]) q.push(v);
}
}
for(int i=0;i<len;i++){
int u=Component[c][i];
if(deg[u]) return false;
}
return true;
}

int main(){
freopen("in.txt","r",stdin);
while(scanf("%d%d",&N,&M)!=EOF){
int u,v;
for(int i=0;i<=N;i++){
V[i].clear();
G[i].clear();
Component[i].clear();
}
for(int i=0;i<M;i++){
scanf("%d%d",&u,&v);
deg[v]++;
V[u].push_back(v);
V[v].push_back(u);
G[u].push_back(v);
}
idx=0;
memset(belong,-1,sizeof(belong));
memset(cnt,0,sizeof(cnt));
for(int i=1;i<=N;i++) if(belong[i]==-1){
DFS(i);
idx++;
}
int ans=0;
for(int i=1;i<=N;i++){
cnt[belong[i]]++;
Component[belong[i]].push_back(i);
}
for(int i=0;i<idx;i++){
if(DAG(i)) ans+=cnt[i]-1;
else ans+=cnt[i];
}
printf("%d\n",ans);
}
return 0;
}