제출 #301194

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
301194		mode149256	친구 (IOI14_friend)	C++17	100 / 100	39 ms	2424 KiB

friend

/*input

*/
#include <bits/stdc++.h>
#include "friend.h"
using namespace std;

namespace my_template {
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}
}
using namespace my_template;

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;


int N;
vi p;
vi q;

int findSample(int n, int confidence[], int host[], int protocol[]) {
	N = n;
	p.resize(N, 0); for (int i = 0; i < N; i++) p[i] = confidence[i];
	q.resize(N, 0);

	for (int y = N - 1; y >= 1; --y)
	{
		int x = host[y];
		if (protocol[y] == 0) {
			p[x] = max(p[x], p[x] + q[y]);
			q[x] = max({q[x], q[x] + q[y], q[x] + p[y]});
		} else if (protocol[y] == 1) {
			p[x] = max({p[x], p[x] + p[y], p[x] + q[y], q[x] + p[y]});
			q[x] = max(q[x], q[x] + q[y]);
		} else if (protocol[y] == 2) {
			p[x] = max({p[x], p[x] + q[y], q[x] + p[y]});
			q[x] = max(q[x], q[x] + q[y]);
		}
	}

	return max(p[0], q[0]);
}

• Subtask 1: 11 / 11
• Subtask 2: 8 / 8
• Subtask 3: 8 / 8
• Subtask 4: 19 / 19
• Subtask 5: 23 / 23
• Subtask 6: 31 / 31