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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
#276203#4815. Flower's LandGeZhiyuanTL 1ms8536kbC++141.9kb2023-12-05 18:04:572023-12-05 18:04:58

Judging History

你现在查看的是最新测评结果

  • [2023-12-05 18:04:58]
  • 评测
  • 测评结果:TL
  • 用时:1ms
  • 内存:8536kb
  • [2023-12-05 18:04:57]
  • 提交

answer

#include<bits/stdc++.h>

using namespace std;

const int N = 4e4 + 5, K = 3000 + 5;
int n = 0, k = 0, tag[N] = {}, a[N] = {}, f[N][K] = {}, g[N][K] = {}, ans[N] = {};
int m = 0, siz[N] = {};
vector<vector<int> > G(N);
vector<int> V;

inline void dfs1(int u, int p = 0){
	siz[u] = 1, m ++;
	for(int v : G[u]) if(!tag[v] && v != p){
		dfs1(v, u);
		siz[u] += siz[v];
	}
}

inline int dfs2(int u, int p = 0){
	for(int v : G[u]) if(!tag[v] && v != p) if(2 * siz[v] > m) return dfs2(v, u);
	return u;
}

inline void dfs3(int u, int p = 0){
	siz[u] = 1, V.push_back(u);
	for(int v : G[u]) if(!tag[v] && v != p){
		dfs3(v, u);
		siz[u] += siz[v];
	}
}

inline void calc(){
	for(int i = 0 ; i <= m ; i ++) memset(f[i], 0xa0, sizeof(f[i])), memset(g[i], 0xa0, sizeof(g[i]));
	f[0][0] = 0, g[m][0] = 0;
	for(int i = 0 ; i < m ; i ++){
		int s = siz[V[i]], w = a[V[i]];
		for(int x = 0 ; x <= k ; x ++){
			f[i + 1][x + 1] = max(f[i + 1][x + 1], f[i][x] + w);
			f[i + s][x] = max(f[i + s][x], f[i][x]);
		}
	}
	for(int i = m - 1 ; i >= 0 ; i --){
		int s = siz[V[i]], w = a[V[i]];
		for(int x = 0 ; x <= k ; x ++){
			g[i][x + 1] = max(g[i][x + 1], g[i + 1][x] + w);
			g[i][x] = max(g[i][x], g[i + s][x]);
		}
	}
	for(int i = 0 ; i < m ; i ++) for(int x = 0 ; x < k ; x ++) if(f[i][x] >= 0 && g[i + 1][k - 1 - x] >= 0)ans[V[i]] = max(ans[V[i]], f[i][x] + g[i + 1][k - 1 - x] + a[V[i]]);
}

inline void cdq(int r){
	m = 0, dfs1(r);
	if(m < k) return;
	r = dfs2(r);
	V.clear(), dfs3(r);
	calc();
	tag[r] = 1; for(int v : G[r]) cdq(v);
}

int main(){
	memset(ans, 0xa0, sizeof(ans));
	scanf("%d %d", &n, &k);
	for(int u = 1 ; u <= n ; u ++) scanf("%d", &a[u]);
	for(int i = 1, u = 0, v = 0 ; i < n ; i ++){
		scanf("%d %d", &u, &v);
		G[u].push_back(v), G[v].push_back(u);
	}
	cdq(1);
	for(int u = 1 ; u <= n ; u ++) printf("%d\n", ans[u]);
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 6676kb

input:

5 3
6 10 4 3 4
3 4
4 2
2 5
5 1

output:

20
20
17
17
20

result:

ok 5 number(s): "20 20 17 17 20"

Test #2:

score: 0
Accepted
time: 1ms
memory: 6492kb

input:

7 4
1 3 2 1 7 12 17
4 6
1 4
5 1
2 5
7 6
3 2

output:

31
13
13
31
21
31
31

result:

ok 7 numbers

Test #3:

score: 0
Accepted
time: 1ms
memory: 6648kb

input:

1 1
20

output:

20

result:

ok 1 number(s): "20"

Test #4:

score: 0
Accepted
time: 0ms
memory: 8536kb

input:

10 3
19 7 25 18 93 97 21 51 60 80
6 7
7 1
1 9
9 10
10 2
2 5
5 3
3 8
8 4

output:

159
180
169
94
180
137
137
169
159
180

result:

ok 10 numbers

Test #5:

score: -100
Time Limit Exceeded

input:

20 3
932 609 248 720 831 253 418 482 1000 542 436 304 217 163 872 380 704 845 497 610
17 12
1 17
15 17
13 17
2 15
16 2
18 16
8 16
4 16
19 4
6 4
20 19
10 19
9 10
5 10
7 9
3 9
14 5
11 7

output:


result: