This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
/*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] = max(q[x], q[x] + p[y]);
} else if (protocol[y] == 1) {
p[x] = max(p[x], p[x] + p[y]);
p[x] = max(p[x], p[x] + q[y]);
p[x] = max(p[x], 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]);
p[x] = max(p[x], q[x] + p[y]);
q[x] = max(q[x], q[x] + q[y]);
}
}
return max(p[0], q[0]);
}
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |