E - Preorder Test

思路:想到二分答案了之后就不难啦, 对于每个答案用树形dp取check, 如果二分的值是val, dp[ i ]表示 i 这棵子树答案不低于val的可以访问的

最多节点, 第二次dfs求出以每个点为根的答案。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define PLI pair<LL, int>
#define ull unsigned long long
using namespace std; const int N = 2e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-; int n, k, a[N], dp[N], sz[N], mx[N], mx2[N];
vector<int> edge[N]; void dfs1(int u, int fa, int val) {
sz[u] = ;
mx[u] = , mx2[u] = , dp[u] = ;
for(int v : edge[u]) {
if(v == fa) continue;
dfs1(v, u, val);
sz[u] += sz[v];
if(dp[v] == sz[v]) dp[u] += dp[v];
else {
if(dp[v] >= mx[u]) mx2[u] = mx[u], mx[u] = dp[v];
else if(dp[v] > mx2[u]) mx2[u] = dp[v];
}
}
dp[u] += mx[u];
if(a[u] < val) dp[u] = ;
}
void dfs2(int u, int fa, int cnt, int val, int &ans) {
if(!dp[u]) {
for(int v : edge[u]) {
if(v == fa) continue;
dfs2(v, u, , val, ans);
}
} else {
int ret = dp[u];
if(cnt == n - sz[u]) ret = max(ret, dp[u] + cnt);
else if(cnt > mx[u]) ret = max(ret, dp[u] - mx[u] + cnt), mx2[u] = mx[u], mx[u] = cnt;
else if(cnt > mx2[u]) mx2[u] = cnt;
ans = max(ans, ret);
for(int v : edge[u]) {
if(v == fa) continue;
if(dp[v] == sz[v]) dfs2(v, u, ret-dp[v], val, ans);
else if(dp[v] == mx[u]) dfs2(v, u, ret-mx[u]+mx2[u], val, ans);
else dfs2(v, u, ret, val, ans);
}
}
} bool check(int val) {
dfs1(, , val);
int ans = ;
dfs2(, , , val, ans);
return ans >= k;
} int main() {
scanf("%d%d", &n, &k);
for(int i = ; i <= n; i++) scanf("%d", &a[i]);
for(int i = ; i < n; i++) {
int u, v; scanf("%d%d", &u, &v);
edge[u].push_back(v);
edge[v].push_back(u);
}
int low = , high = , ans = ;
while(low <= high) {
int mid = low + high >> ;
if(check(mid)) ans = mid, low = mid + ;
else high = mid - ;
}
printf("%d\n", ans);
return ;
} /*
*/

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