| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 976683 | josanneo22 | Road Closures (APIO21_roads) | C++17 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "roads.h"
#include <vector>
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
#define vi vector<int>
vector<i64> N2(int N, vi U, vi V, vi W) {
vector<vector<pair<int, int>>> G(N);
for (int i = 0; i < N - 1; i++) {
G[U[i]].push_back(make_pair(V[i], W[i]));
G[V[i]].push_back(make_pair(U[i], W[i]));
}
auto solve = [&](int k) {
// cout << k << '\n';
vector<vector<i64>> dp(N, vector<i64>(2, 100000));
function<void(int, int)> dfs = [&](int u, int f) {
// cout << "tree : " << u << ' ' << f << '\n';
// if (G[u].size() < k) {
// dp[u][0] = dp[u][1] = 0;
// }
i64 sum = 0;
vector<i64> wait;
for (auto & v : G[u]) {
if (v.first == f) continue;
dfs(v.first, u);
wait.push_back(-dp[v.first][1] + dp[v.first][0] + v.second);
sum += dp[v.first][0] + v.second;
// cout << v.first << ' ' << dp[v.first][0] + v.second << ' ' << dp[v.first][1] << '\n';
}
// cout << "total: " << sum << '\n';
sort(wait.begin(), wait.end(), greater<i64>());
// cout << "wait : ";
// for (auto & v : wait) cout << v << ' ';
// cout << '\n';
int deg = G[u].size();
int remain = deg - max(deg - k - 1, 0);
// cout << "remain: " << remain << '\n';
i64 res = sum, S = sum;
for (int i = 0; i < min(remain, (int)wait.size()); i++) {
S -= wait[i];
res = min(res, S);
}
dp[u][0] = res;
remain = deg - max(deg - k, 0);
res = sum; S = sum;
// cout << "remain: " << remain << '\n';
// cout << "sums : ";
for (int i = 0; i < min(remain, (int)wait.size()); i++) {
S -= wait[i];
res = min(res, S);
}
// cout << '\n';
dp[u][1] = res;
// cout << "++++++++++++++\n";
};
dfs(0, -1);
// for (int i = 0; i < N; i++) {
// cout << "dp : " << i << ' ' << dp[i][0] << ' ' << dp[i][1] << '\n';
// }
return dp[0][0];
};
vector<i64> ans(N);
for (int k = 0; k < N; k++) {
ans[k] = solve(k - 1);
}
return ans;
}
vector<i64> juhua(int N, vi U, vi V, vi W) {
vector<i64> ans(N);
ans[0] = accumulate(W.begin(), W.end(), 0LL);
sort(W.begin(), W.end(), greater<int>());
for (int i = 1; i < N; i++) {
ans[i] = ans[i - 1] - W[i - 1];
}
return ans;
}
vector<i64> chain(int N, vi U, vi V, vi W) {
vector<i64> ans(N);
ans[0] = accumulate(W.begin(), W.end(), 0LL);
vector<i64> dp(N);
for (int i = 0; i < N - 1; i++) {
i64 take = (i == 0 ? 0 : dp[i - 1]) + W[i];
i64 dun_take = (i - 2 < 0 ? 0 : dp[i - 2]) + (i - 1 < 0 ? 0 : W[i - 1]);
dp[i] = min(take, dun_take);
}
ans[1] = dp[N - 2];
return ans;
}
vector<i64> minimum_closure_costs(int N, vi U, vi V, vi W) {
if (N <= 5000) return N2(N, U, V, W);
if (count(U.begin(), U.end(), 0) == N - 1) return juhua(N, U, V, W);
return chain(N, U, V, W);
}
#include "grader.cpp"
/*
ST1: 菊花图
直接一个一个贪心移除
ST2: 链
我们可以直接dp, 从i - 1/i - 2转移过来
ST3 & 4:
对于每个K, 我们可以单独solve
就是说如果我们让dp[i] = 整个i的子树里最小价格满足, deg[i] <= k
发现到其实这样的存储方式不足以满足dp转移方式
dp[u][0/1] = 和父亲节点是否有边
如果我们拿这个边那么我们会得到dp[v][1]
如果我们不要这个边我们得到dp[v][0] + W[i]
如果我们现在和父亲节点连边: 需要移除min(deg[u] - k, 0)
没和父亲连边： 需要移除min(deg[u] - k - 1, 0)
那么我们可以拿的边 = deg[u] - 移除
我们先把dp[v][1] - (dp[v][0] + W[i])排序， ans = sum(dp[v][0] + W[i])
然后我们降序排序然后直接贪心取
O(N ^ 2) 24分
若W[i] = 1, 就是说我们只是求最少移除多少遍使得
max deg <= k
erm....
有没有可能是贪心？
猜测: 如果现在这个边的deg > k, 移除是最优的
然后就是贪心
打了数据，发现不对
*/