이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "friend.h"
#include <bits/stdc++.h>
#define eb emplace_back
#define pb push_back
#define ft first
#define sd second
#define pi pair<int, int>
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define dbg(...) dbg_out(__VA_ARGS__)
using ll = long long;
using ld = long double;
using namespace std;
//Constants
const ll INF = 5 * 1e18;
const int IINF = 2 * 1e9;
const ll MOD = 1e9 + 7;
// const ll MOD = 998244353;
const ll dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const ld PI = 3.14159265359;
template<typename T> void mins(T& x, T y) {x = min(x, y);}
template<typename T> void maxs(T& x, T y) {x = max(x, y);}
vector<vector<int>> e;
vector<int> a;
int subtask1() {
int n = a.size();
int ans = 0;
for (int mask = 0; mask < (1 << n); mask++) {
bool ok = true;
int cur = 0;
for (int i = 0; i < n; i++) {
if (mask & (1 << i)) {
cur += a[i];
for (int to : e[i]) {
if (mask & (1 << to)) {
ok = false;
}
}
}
}
if (ok) maxs(ans, cur);
}
return ans;
}
int subtask2() {
int ans = 0;
for (int x : a) {
ans += x;
}
return ans;
}
int subtask3() {
int ans = 0;
for (int x : a) {
maxs(ans, x);
}
return ans;
}
int subtask4() {
int ans = 0;
int n = a.size();
vector<bool> vis(n + 5, false);
vector<vector<int>> dp(n + 5, vector<int> (2, 0));
function<void(int)> dfs = [&](int f) {
vis[f] = true;
dp[f][1] += a[f];
for (int to : e[f]) {
if (!vis[to]) {
dfs(to);
dp[f][0] += max(dp[to][1], dp[to][0]);
dp[f][1] += dp[to][0];
}
}
};
for (int i = 0; i < n; i++) {
if (!vis[i])
ans += max(dp[i][0], dp[i][1]);
}
return ans;
}
// Find out best sample
int findSample (int n, int confidence[], int host[],int protocol[]){
e.resize(n + 5); a.resize(n);
for (int i = 0; i < n; i++)
a[i] = confidence[i];
bool sub1, sub2, sub3, sub4;
sub1 = sub2 = sub3 = sub4 = true;
for (int i = 1; i < n; i++) {
sub2 &= protocol[i] == 1;
sub3 &= protocol[i] == 2;
sub4 &= protocol[i] == 0;
if (protocol[i] == 0) {
e[host[i]].pb(i);
e[i].pb(host[i]);
} else if (protocol[i] == 1) {
for (int to : e[host[i]]) {
e[to].pb(i);
e[i].pb(to);
}
} else {
for (int to : e[host[i]]) {
e[to].pb(i);
e[i].pb(to);
}
e[host[i]].pb(i);
e[i].pb(host[i]);
}
}
sub1 &= n <= 10;
if (sub1) return subtask1();
else if (sub2) return subtask2();
else if (sub3) return subtask3();
else if (sub4) return subtask4();
return 0;
}
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