Title description
Every cow has a dream: to become the most popular cow in a group! In a herd of N(1<=N<=10,000) cows, give you M(1<=M<=50,000) two-tuples (A, B) to indicate that A thinks B is popular. Since popularity is transferable, if A thinks B is popular, and B thinks C is popular, A will also think C is popular, even if this is not a very clear rule. Your task is to count the number of cows that are liked by all other cows.
enter
In the first line, two numbers, N and M. Lines 2～M+1 have two numbers in each line, A and B, indicating that A thinks B is popular.
Output
A number, the number of cows considered popular by all other cows.
Input sample
3
1 2
2 1
2 3
Sample output
1
Sample description
Cow No. 3 is the only one considered famous by all other cows.
Description
Data range limit
1 <= N <= 10, 000 1<=N<=10,0001<=N<=10,000
1 < = M < = 50 , 000 1<=M<=50,000 1<=M<=50,000
analysis
O (n 2) O (n ^ 2) O ( n2 )Over a million who can be me! ?
Connect the edges directly to the adjacent table, and then search for each cow linked to it. If the number reaches n-1, it means that ta is a POPULAR cow.
Then add it to the ans array, and finally count the ans array
Upload code
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m,v[100001],ans[100001],s;
int h[100001],tot;
struct node
{
int x,y;
}e[100001];
void add(int x,int y)
{
tot++;
e[tot].x=y;
e[tot].y=h[x];
h[x]=tot;
}
void dfs(int x)
{
v[x]=1;
for(int i=h[x];i>0;i=e[i].y)
{
int t=e[i].x;
if(!v[t])
{
ans[t]++;
dfs(t);
}
}
}
int main()
{
cin>>n>>m;
for(int i=1;i<=m;i++)
{
int a,b;
cin>>a>>b;
add(a,b);
}
for(int i=1;i<=n;i++)
{
memset(v,0,sizeof(v));
dfs(i);
}
for(int i=1;i<=n;i++)
{
if(ans[i]==n-1) s++;
}
cout<<s;
return 0;
}
Finished this question smoothly AK.