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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#628299 | #6438. Crystalfly | MahnoKropotkinov# | WA | 1ms | 3844kb | C++20 | 2.9kb | 2024-10-10 19:32:58 | 2024-10-10 19:32:58 |
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, psum = pMax, ssum = 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[j];
ssum[j] = ssum[j + 1] + dp[nxt];
sMax[j] = max(sMax[j + 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[j - 1], sMax[j + 1] + a[nxt]));
}
};
dfs(1, 0);
cout << dp[1] + a[1] << '\n';
// cout<<max(dp[1][])
}();
}
Details
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Test #1:
score: 0
Wrong Answer
time: 1ms
memory: 3844kb
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 10101
result:
wrong answer 2nd numbers differ - expected: '10111', found: '10101'