QOJ.ac
QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#749550 | #2562. Fake Plastic Trees 2 | arvindr9 | WA | 0ms | 3576kb | C++20 | 5.7kb | 2024-11-15 03:42:48 | 2024-11-15 03:42:49 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define DEBUG(...) debug(#__VA_ARGS__, __VA_ARGS__)
#else
#define DEBUG(...) 6
#endif
template<typename T, typename S> ostream& operator << (ostream &os, const pair<T, S> &p) {return os << "(" << p.first << ", " << p.second << ")";}
template<typename C, typename T = decay<decltype(*begin(declval<C>()))>, typename enable_if<!is_same<C, string>::value>::type* = nullptr>
ostream& operator << (ostream &os, const C &c) {bool f = true; os << "["; for (const auto &x : c) {if (!f) os << ", "; f = false; os << x;} return os << "]";}
// Add this operator for __int128
ostream& operator << (ostream &os, __int128 x) {
if (x < 0) {
os << '-';
x = -x;
}
if (x > 9) os << (x / 10);
os << (int)(x % 10);
return os;
}
template<typename T> void debug(string s, T x) {cerr << "\033[1;35m" << s << "\033[0;32m = \033[33m" << x << "\033[0m\n";}
template<typename T, typename... Args> void debug(string s, T x, Args... args) {for (int i=0, b=0; i<(int)s.size(); i++) if (s[i] == '(' || s[i] == '{') b++; else
if (s[i] == ')' || s[i] == '}') b--; else if (s[i] == ',' && b == 0) {cerr << "\033[1;35m" << s.substr(0, i) << "\033[0;32m = \033[33m" << x << "\033[31m | "; debug(s.substr(s.find_first_not_of(' ', i + 1)), args...); break;}}
typedef int int2;
#define int __int128
#define pi pair<int, int>
#define vi vector<int>
#define vii vector<vector<int>>
#define vpi vector<pi>
#define lep(i,l,r) for(int i=l;i<=r;++i)
#define rep(i,r,l) for(int i=r;i>=l;--i)
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define f first
#define s second
long long t;
const int maxn = 1005;
const int maxk = 51;
vector<int> dp[maxn][maxk];
int2 main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> t;
while (t--) {
int n, k, l, r;
{
long long a, b, c, d;
cin >> a >> b >> c >> d;
n = a;
k = b;
l = c;
r = d;
}
vector<int> a(n+1);
lep(i,1,n) {
long long x;
cin >> x;
a[i] = x;
}
vector<vector<int>> adj(n+1);
lep(i,1,n-1) {
long long u,v;
cin >> u >> v;
adj[u].pb(v);
adj[v].pb(u);
}
vector<int> sub(n+1);
vector<int> num_sub(n+1);
auto dfs1 = [&](auto &self, int u, int p) -> void {
int val = a[u];
int num = 1;
for (int v: adj[u]) {
if (v != p) {
self(self,v,u);
val += sub[v];
num += num_sub[v];
}
}
sub[u] = val;
num_sub[u] = num;
// assert(sub[u] >= 0);
};
dfs1(dfs1,1,0);
auto simplify = [&](vector<int> a) -> vector<int> {
if (a.empty()) return a;
sort(a.begin(),a.end());
a.erase(unique(a.begin(),a.end()),a.end());
int n = a.size();
vector<int> res = {a[0]};
if (a.size() == 1) return res;
lep(i,2,n-1) {
if (a[i] <= res.back() + (r-l)) {
// don't push earlier element
continue;
} else {
res.pb(a[i-1]);
}
}
res.pb(a.back());
return res;
};
auto dfs2 = [&](auto &self, int u, int p) -> void {
int sz = num_sub[u];
vector<vector<int>> curdp(min(k+2,sz+1),vector<int>());
curdp[0].pb(0);
for (int v: adj[u]) {
if (v != p) {
self(self,v,u);
vector<vector<int>> nextdp(min(k+2,sz+1),vector<int>());
int szv = min(k,num_sub[v]);
lep(k1,0,sz) {
lep(k2,0,szv) {
for (int x1: curdp[k1]) {
for (int x2: dp[v][k2]) {
if (k1 + k2 <= k+1 and x1 + x2 <= r) {
nextdp[k1+k2].pb(x1+x2);
}
}
}
}
}
lep(i,0,min(k+1,sz)) {
nextdp[i] = simplify(nextdp[i]);
}
swap(curdp, nextdp);
}
}
// add u
DEBUG(u);
DEBUG(curdp);
lep(i,0,min(k+1,sz)) {
for (int x: curdp[i]) {
int next_val = x + a[u];
if (next_val <= r) {
dp[u][i].pb(next_val);
}
if (next_val >= l and next_val <= r and i < k + 1) {
dp[u][i+1].pb(0);
}
}
simplify(dp[u][i]);
DEBUG(i,dp[u][i]);
}
};
dfs2(dfs2,1,0);
lep(i,1,k+1) {
if (!dp[1][i].empty() and dp[1][i][0] == 0) {
cout << 1;
} else {
cout << 0;
}
}
cout << "\n";
lep(i,1,n) {
lep(j,0,k) {
dp[i][j].clear();
}
}
}
}
/* stuff you should look for
* int overflow, array bounds
* special cases (n=1?)
* do smth instead of nothing and stay organized
* WRITE STUFF DOWN
* DON'T GET STUCK ON ONE APPROACH
*/
/*
4 3 1 2
1 1 1 1
1 2
2 3
3 4
*/
詳細信息
Test #1:
score: 0
Wrong Answer
time: 0ms
memory: 3576kb
input:
3 4 3 1 2 1 1 1 1 1 2 2 3 3 4 4 3 1 2 1 1 1 1 1 2 1 3 1 4 4 3 0 0 1 1 1 1 1 2 1 3 1 4
output:
0111 0011 0001
result:
wrong answer 3rd words differ - expected: '0000', found: '0001'