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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#628318 | #6438. Crystalfly | MahnoKropotkinov# | WA | 128ms | 3584kb | C++20 | 2.9kb | 2024-10-10 19:36:59 | 2024-10-10 19:36:59 |
Judging History
answer
#include <algorithm>
#include <array>
#include <functional>
#include <iostream>
#include <vector>
using namespace std;
using ll = long long;
int calc2(int i, int j) { return i + j + 2 > 4 ? 4 : i + j + 2; }
int calc1(int i, int j) { return i + j + 1 > 4 ? 4 : i + j + 1; }
// #define debug
const int maxt = 4 + 1;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(0);
int t;
cin >> t;
while (t--)
[]() {
int n;
cin >> n;
auto a = vector(n + 1, 0);
auto t = vector(n + 1, 0);
for (int i = 1; i <= n; i++)
cin >> a[i];
for (int i = 1; i <= n; i++)
cin >> t[i];
// auto dp = vector(n + 1, array<array<array<ll, 2>, 2>, maxt>{}); // 第四个维度表示 是否已经选择到了那个一去不回的点
// auto buf = array<array<array<ll, 2>, 2>, maxt>{};
auto dp = vector(n + 1, 0ll);
auto edge = vector(n + 1, vector(0, 0));
for (int i = 1; i < n; i++) {
int u, v;
cin >> u >> v;
edge[u].push_back(v);
edge[v].push_back(u);
}
function<void(int, int)> dfs = [&](int cur, int pre) {
// dp[cur][0][1][1] = dp[cur][0][0][1] = 0; // arrive there and don't go to any node of subtree
int Max = 0;
ll sum = 0;
for (int nxt : edge[cur]) {
if (nxt == pre)
continue;
dfs(nxt, cur);
Max = max(Max, a[nxt]);
dp[cur] += dp[nxt];
sum += dp[nxt];
}
dp[cur] += Max; // 走子节点 只走其中一个
auto pMax = vector(edge[cur].size() + 2, -__LONG_LONG_MAX__), sMax = pMax, id = pMax;
int j = 0;
for (int i = 1; i <= edge[cur].size(); i++) {
int nxt = edge[cur][i - 1];
if (nxt == pre)
continue;
++j;
id[j] = nxt;
// psum[j] = psum[j - 1] + dp[nxt];
pMax[j] = max(pMax[j - 1], a[nxt] - dp[nxt]);
}
for (int k = j; k >= 1; --k) {
int nxt = id[k];
// ssum[j] = ssum[j + 1] + dp[nxt];
sMax[k] = max(sMax[k + 1], a[nxt] - dp[nxt]);
}
for (int k = 1; k <= j; k++) {
int nxt = id[k];
if (t[nxt] == 3)
dp[cur] = max(dp[cur], sum + max(pMax[k - 1], sMax[k + 1]) + a[nxt]);
}
};
dfs(1, 0);
cout << dp[1] + a[1] << '\n';
// cout<<max(dp[1][])
}();
}
Details
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Test #1:
score: 100
Accepted
time: 0ms
memory: 3584kb
input:
2 5 1 10 100 1000 10000 1 2 1 1 1 1 2 1 3 2 4 2 5 5 1 10 100 1000 10000 1 3 1 1 1 1 2 1 3 2 4 2 5
output:
10101 10111
result:
ok 2 number(s): "10101 10111"
Test #2:
score: 0
Accepted
time: 0ms
memory: 3548kb
input:
10 6 8 1 1 5 8 9 2 1 2 2 2 2 1 2 2 3 2 4 1 5 2 6 6 6 4 4 1 3 6 2 1 3 3 3 3 1 2 1 3 3 4 4 5 5 6 6 10 5 1 8 5 1 1 3 1 2 2 2 1 2 2 3 2 4 2 5 3 6 10 6 8 8 9 6 9 5 6 6 4 2 1 3 3 2 2 2 2 3 1 1 2 1 3 3 4 4 5 5 6 4 7 2 8 7 9 9 10 7 10 9 1 5 7 5 4 1 1 1 2 1 3 2 1 2 1 3 3 4 3 5 5 6 1 7 5 7 1 1 4 2 3 1 3 2 2 1...
output:
25 24 24 56 31 14 16 28 19 19
result:
ok 10 numbers
Test #3:
score: -100
Wrong Answer
time: 128ms
memory: 3564kb
input:
100000 10 9 1 7 9 5 1 10 5 3 8 2 1 1 3 1 2 2 3 3 1 1 2 2 3 3 4 1 5 2 6 2 7 6 8 7 9 7 10 3 6 6 1 2 1 2 1 2 1 3 10 6 5 3 7 1 5 1 9 7 3 3 1 3 3 1 3 2 2 2 3 1 2 1 3 3 4 4 5 2 6 6 7 4 8 7 9 1 10 7 2 8 9 7 7 9 10 2 3 2 2 3 2 1 1 2 2 3 1 4 3 5 4 6 3 7 1 8 1 1 4 2 7 9 9 9 8 4 2 7 3 1 2 1 1 1 1 1 2 2 3 2 4 3...
output:
49 12 41 45 8 4 38 22 20 21 5 19 23 44 26 5 21 28 28 32 36 15 5 26 38 36 20 35 27 36 20 9 32 32 22 11 41 15 20 54 38 20 45 36 20 29 24 4 37 44 30 44 17 17 36 29 3 6 24 44 25 28 49 13 5 1 44 27 17 21 15 17 17 24 29 39 10 39 38 26 22 24 9 17 41 7 28 33 51 18 14 14 7 35 23 13 11 43 30 24 35 2 43 33 17 ...
result:
wrong answer 52nd numbers differ - expected: '45', found: '44'