Submission #945528

#	Time	Username	Problem	Language	Result	Execution time	Memory
945528		vjudge1	Magic Tree (CEOI19_magictree)	C++17	34 / 100	2045 ms	60076 KiB

magictree

#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
 
#define ff first
#define ss second
#define pb push_back
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define int long long
#define rnd(l, r) uniform_int_distribution<int>(l, r)(rng)
 
using namespace std;
 
void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);} 
int pow(int a,int b,int m){int ans=1;while(b){if(b&1){ans=(ans*a)%m;}b>>=1;a=(a*a)%m;}return ans;}
int binpow(int a,int b){int ans=1;while(b){if(b&1){ans=(ans*a);}b>>=1;a=(a*a);}return ans;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
const int N = 1e5 + 100, mxc = 105, inf = 1e18, MX = 1e5 + 10;

int cost[N][25], dp[N][25];

vector <int> g[N];

void setmax(int &a, int b){
	a = max(a, b);
}

void dfs(int v = 1, int par = 1, int time = 1){
	dp[v][time] = max(dp[v][time - 1], cost[v][time]);
	int sum = 0;
	for(auto to : g[v]){
		if(to == par)continue;
		dfs(to, v, time);
		sum += dp[to][time];
		dp[v][time] = max(dp[v][time], dp[to][time] + cost[v][time]);
	}
	dp[v][time] = max(dp[v][time], sum + cost[v][time]);
	//setmax(dp[v][time], sum + cost[v][time]);
	//cout << v <<"-"<< time <<" "<< dp[v][time] << endl;
}
main(){
	iostream::sync_with_stdio(false);	
    cin.tie(nullptr);
    cout.tie(nullptr);
    int n, m, k;
    cin >> n >> m >> k;
    for(int i = 2; i <= n; i++){
		int a;
		cin >> a;
		g[a].pb(i);
		g[i].pb(a);
	}
	
	for(int i = 1 ; i<= m; i++){
		int a, b, c;
		cin >> a >> b >> c;
		cost[a][b] = c;
	}
	
	for(int i = 1; i <= k; i++){
		dfs(1, 1, i);
	}
	cout << dp[1][k] << endl;
}	
/*
 * Before implementing the problem:
	
	Do I understand the problem correctly?
	Which places are tricky? What do I need to remember to implement them correctly?
	Which place is the heaviest by implementation? Can I do it simpler?
	Which place is the slowest? Where do I need to be careful about constant factors and where I can choose slower but simpler implementation?
	----------------------------------
	If not AC:
 
	Did you remember to do everything to handle the tricky places you thought about before?
	Is the solution correct?
	Do I understand the problem correctly?
	----------------------------------
	If you have a test on which the solution gives wrong answer:
 
	Is the solution doing what it was supposed to do?
	Is the problem in the code or in the idea?
*/

Compilation message (stderr)

magictree.cpp:43:1: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
   43 | main(){
      | ^~~~
magictree.cpp: In function 'void fp(std::string)':
magictree.cpp:15:29: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   15 | void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);}
      |                      ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
magictree.cpp:15:70: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   15 | void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);}
      |                                                               ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 6 / 6
• Subtask 2: 0 / 3
• Subtask 3: 0 / 11
• Subtask 4: 12 / 12
• Subtask 5: 16 / 16
• Subtask 6: 0 / 13
• Subtask 7: 0 / 22
• Subtask 8: 0 / 17