Cfdiv1+2-pathwalks- (tree array + linear DP)

F

The question ：
Is to give you a n spot m edge , Then let you choose a path , Then the order of the sides you walk , It needs to be in the order of increasing , Then the edge weight should also increase .

reflection ：
At first, I felt that such a problem was graph theory torture problem , But I have found so many people , Found that this is a dp. Since the order of the edges to be given , Then it must be enumerating edges in order , Before and after , Definition dp[i] Choose the first i The maximum value of the edge . Then come to the i One side , Or start from this side , Or transfer from the front side , Transfer from the edge whose end point is the starting point of this edge , At the same time, ensure the weight of that side < Current weight . If violence starts from vector Look inside , This is a m*m Complexity . So every query needs log(m), And the weight is smaller than the current , Find a maximum . Then maintain the tree array , But what I found was that , In this way, we should establish n A tree , This space is too complicated . But I think it can be used map Build up trees , Then the actual space will not be so large , Because every time +bit(x), Not many points are actually occupied . Actually, I thought map Build up trees , It's still the time when we used to do and search collections , Need to maintain a large point , You can use it map When acc Array is ok .

Code ：

#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define db double
#define int long long
#define PII pair<int,int >
#define mem(a,b) memset(a,b,sizeof(a))
#define IOS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
		
using namespace std;
const int mod = 1e9+7,inf = 1e18;
const int N = 2e5+10,M = 2010;

int T,n,m,k;
PII va[N];
int vb[N];
int dp[N];

int R = 1e5+5;
map<int,int > tr[N];

int bit(int x)
{
    
	return x&(-x);
}

void update(int x,int value,int op)
{
    
	while(x<=R)
	{
    
		tr[op][x] = max(tr[op][x],value);
		x += bit(x);
	}
}

int query(int x,int op)
{
    
	int maxn = 0;
	while(x)
	{
    
		maxn = max(maxn,tr[op][x]);
		x -= bit(x);
	}
	return maxn;
}

signed main()
{
    
	IOS;
	cin>>n>>m;
	for(int i=1;i<=m;i++)
	{
    
		cin>>va[i].fi>>va[i].se>>vb[i];
		vb[i] += 1; // It is only related to the size of the weight ,+1 To prevent the following query( negative ) and update(0)
	}
	for(int i=1;i<=m;i++)
	{
    
		int a = va[i].fi,b = va[i].se,c = vb[i];
		dp[i] = max(1ll,query(c-1,a)+1);
		update(c,dp[i],b);
	}
	int maxn = 1;
	for(int i=1;i<=m;i++) maxn = max(maxn,dp[i]);
	cout<<maxn;
	return 0;
}

summary ：
Think more , Try more .