CodeForces - 459E - Pashmak and Graph

先上题目：

E. Pashmak and Graph

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Pashmak's homework is a problem about graphs. Although he always tries to do his homework completely, he can't solve this problem. As you know, he's really weak at graph theory; so try to help him in solving the problem.

You are given a weighted directed graph with n vertices and m edges. You need to find a path (perhaps, non-simple) with maximum number of edges, such that the weights of the edges increase along the path. In other words, each edge of the path must have strictly greater weight than the previous edge in the path.

Help Pashmak, print the number of edges in the required path.

Input
The first line contains two integers n, m (2 ≤ n ≤ 3·105; 1 ≤ m ≤ min(n·(n - 1), 3·105)). Then, m lines follows. The i-th line contains three space separated integers: ui, vi, wi (1 ≤ ui, vi ≤ n; 1 ≤ wi ≤ 105) which indicates that there's a directed edge with weight wi from vertex ui to vertex vi.

It's guaranteed that the graph doesn't contain self-loops and multiple edges.

Output
Print a single integer — the answer to the problem.

Sample test(s)

input

3 3
1 2 1
2 3 1
3 1 1

output

1

input

3 3
1 2 1
2 3 2
3 1 3

output

3

input

6 7
1 2 1
3 2 5
2 4 2
2 5 2
2 6 9
5 4 3
4 3 4

output

6

Note
In the first sample the maximum trail can be any of this trails:

#include <bits/stdc++.h>
#define MAX 300002
#define ll long long
using namespace std;

typedef struct edge{
ll u,v,w;

bool operator < (const edge& o)const{
return w<o.w;
}
}edge;
edge e[MAX];
int n,m,maxn;
int dp[MAX],f[MAX];

int main()
{
//freopen("data.txt","r",stdin);
ios::sync_with_stdio(false);
while(cin>>n>>m){
memset(dp,0,sizeof(dp));
memset(f,0,sizeof(f));
for(int i=0;i<m;i++){
cin>>e[i].u>>e[i].v>>e[i].w;
}
sort(e,e+m);
maxn=0;
for(int i=0;i<m;i++){
int j;
for(j=i;j<m;j++) if(e[i].w!=e[j].w) break;
for(int k=i;k<j;k++){
f[e[k].v]=max(dp[e[k].u]+1,f[e[k].v]);
}
for(int k=i;k<j;k++){
dp[e[k].v]=max(f[e[k].v],dp[e[k].v]);
}
i=j-1;
}
for(int i=1;i<=n;i++) maxn=max(dp[i],maxn);
cout<<maxn<<endl;
}
return 0;
}

/*459E*/