Problem:
Score : 100 points

Problem Statement

There is a directed graph G with N vertices and M edges.
The vertices are numbered 1, 2, \ldots, N, and for each i (1 \leq i \leq M), the i-th directed edge goes from Vertex x_i to y_i.
G does not contain directed cycles.

Find the length of the longest directed path in G.
Here, the length of a directed path is the number of edges in it.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq x_i, y_i \leq N

All pairs (x_i, y_i) are distinct.

G does not contain directed cycles.

Input

Input is given from Standard Input in the following format:

N M
x_1 y_1
x_2 y_2
:
x_M y_M

Output

Print the length of the longest directed path in G.

Sample Input 1

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

Sample Output 1

3

The red directed path in the following figure is the longest:

Sample Input 2

6 3
2 3
4 5
5 6

Sample Output 2

2

The red directed path in the following figure is the longest:

Sample Input 3

5 8
5 3
2 3
2 4
5 2
5 1
1 4
4 3
1 3

Sample Output 3

3

The red directed path in the following figure is one of the longest:

Solution in C (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#define MAX_N 100000
typedef struct e{
  int a[1000];
  int point;
  int point2;
}el;
int max(int a,int b){
  if(a>b)return a;
  else return b;
}
int s[MAX_N+1];
el edge[MAX_N+1];
int dp[MAX_N];
int main(){
  int n,m;
  int x,y;
  int i,j;
  int idx = 0;

  scanf("%d %d", &n,&m);
  for(int i=0; i<m;i++){
    scanf("%d %d", &x, &y);
    edge[x-1].a[edge[x-1].point] = y-1;
    edge[x-1].point+=1;
    edge[y-1].point2 += 1;
  }


  for(int i=0; i<n; i++){
    if(edge[i].point2==0)s[idx++]=i;
  }
  j = 0;
  while(j<idx){
    for(int k=0; k<edge[s[j]].point; k++){
      edge[edge[s[j]].a[k]].point2 -= 1;
      if(edge[edge[s[j]].a[k]].point2==0){
        s[idx++] = edge[s[j]].a[k];
      }
    }
    j+=1;
  }
  int ans = 0;
  for(i=n-1;i>-1;i--){
    //printf("%d: ", s[i]);
    for(j=0;j<edge[s[i]].point;j++){
      //printf("%d ", edge[s[i]].a[j]);
      dp[s[i]] = max(dp[s[i]], dp[edge[s[i]].a[j]]+1);
    }
    ans = max(ans, dp[s[i]]);
  }

  printf("%d\n", ans);

  return 0;
}