1 second
256 megabytes
standard input
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.
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.
Print a single integer — the answer to the problem.
3 3 1 2 1 2 3 1 3 1 1
1
3 3 1 2 1 2 3 2 3 1 3
3
6 7 1 2 1 3 2 5 2 4 2 2 5 2 2 6 9 5 4 3 4 3 4
6
In the first sample the maximum trail can be any of this trails: 
.
In the second sample the maximum trail is 
.
In the third sample the maximum trail is 
.
题意是给定n个点m条边，求一条最长路径，使得路径上的边的权值严格递增
这题原来是ccr给的代码，但是实际上还是挺水的……我觉得最难的是A和C啊QAQ
先把边按权值排序，然后令f[i]表示以i结尾的路径的最大长度
然后一条一条加进去就好了
直接贴ccr的代码好了……不想自己再打了
#include<bits/stdtr1c++.h>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
typedef double ld;
typedef struct edge{
int f, t, v;
bool operator<(const edge& e) const{
return v < e.v;
}
}edge;
const int N = 3e5 + 50;
int ans[N];
vector< pi > w;
int main() {
int n, m, a, b, c;
vector< edge > v;
cin >> n >> m;
for(int i = 0; i < m; i++){
cin >> a >> b >> c;
edge e = {a, b, c};
v.push_back(e);
}
sort(v.begin(), v.end());
int i = 0, j;
while( i < v.size()){
j = i;
while(j < v.size() && v[j].v == v[i].v){
if(ans[v[j].f] + 1 > ans[v[j].t]) w.push_back(pi(v[j].t, ans[v[j].f] + 1));
j++;
}
for(int q = 0; q < w.size(); q++){
ans[w[q].first] = max(ans[w[q].first], w[q].second);
}
i = j;
w.clear();
}
cout << *max_element(ans, ans + N) << endl;
}