P6326
Prob
树上选择一些物品,有多个,要求选择的所有物品成一个连通块,求出最大的价值。
Sol
点分治,考虑强制选择某个点
设点分治所有选择的重心按顺序分别为
对于每一个连通块
。 。
显然每个连通块只有且只有一个
证明有误请大神们在评论区留言/kel
Code
#include <bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define pf push_front
#define desktop "C:\\Users\\incra\\Desktop\\"
#define IOS ios :: sync_with_stdio (false),cin.tie (0),cout.tie (0)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair <int,int> PII;
const int dx[] = {1,0,-1,0},dy[] = {0,-1,0,1};
template <typename T1,typename T2> bool tomax (T1 &x,T2 y) {
if (y > x) return x = y,true;
return false;
}
template <typename T1,typename T2> bool tomin (T1 &x,T2 y) {
if (y < x) return x = y,true;
return false;
}
LL power (LL a,LL b,LL p) {
LL ans = 1;
while (b) {
if (b & 1) ans = ans * a % p;
a = a * a % p;
b >>= 1;
}
return ans;
}
int fastio = (IOS,0);
#define endl '\n'
#define puts(s) cout << (s) << endl
const int N = 510,M = 4010;
int n,m;
vector <int> g[N];
int rt;
bool del[N];
int v[N],w[N],d[N];
int ww[N],vv[N];
int ans;
int dfn[N],num[N],last[N],timestamp;
int f[N][M];
int get_size (int u,int fa) {
int s = 1;
for (int v : g[u]) {
if (v == fa || del[v]) continue;
s += get_size (v,u);
}
return s;
}
int get_root (int u,int fa,int tot) {
int maxx = 0,s = 1;
for (int v : g[u]) {
if (v == fa || del[v]) continue;
int t = get_root (v,u,tot);
tomax (maxx,t);
s += t;
}
tomax (maxx,tot - s);
if (maxx <= tot / 2) rt = u;
return s;
}
void DFS (int u,int fa) {
num[dfn[u] = ++timestamp] = u;
for (int v : g[u]) {
if (v == fa || del[v]) continue;
DFS (v,u);
}
last[u] = timestamp;
}
void calc (int u) {
timestamp = 0;
DFS (u,-1);
for (int i = timestamp;i >= 1;i--) {
int s = d[num[i]] - 1;
int idx = 0;
for (int j = 1;j <= s;j <<= 1) ww[++idx] = j * w[num[i]],vv[idx] = j * v[num[i]],s -= j;
if (s) ww[++idx] = s * w[num[i]],vv[idx] = s * v[num[i]];
for (int j = m;j >= w[num[i]];j--) tomax (f[i][j],f[i + 1][j - w[num[i]]] + v[num[i]]);
for (int k = 1;k <= idx;k++) {
for (int j = m;j >= ww[k];j--) tomax (f[i][j],f[i][j - ww[k]] + vv[k]);
}
if (i > 1) {
for (int j = 0;j <= m;j++) tomax (f[i][j],f[last[num[i]] + 1][j]);
}
}
tomax (ans,f[1][m]);
for (int i = 1;i <= timestamp;i++) for (int j = 0;j <= m;j++) f[i][j] = 0;
}
void solve (int _u) {
get_root (_u,-1,get_size (_u,-1));
int u = rt;
calc (u);
del[u] = 1;
for (int v : g[u]) {
if (del[v]) continue;
solve (v);
}
}
void mian () {
cin >> n >> m;
for (int i = 1;i <= n;i++) g[i].clear (),del[i] = 0;
for (int i = 1;i <= n;i++) cin >> v[i];
for (int i = 1;i <= n;i++) cin >> w[i];
for (int i = 1;i <= n;i++) cin >> d[i];
for (int i = 1;i <= n - 1;i++) {
int a,b;
cin >> a >> b;
g[a].pb (b),g[b].pb (a);
}
ans = 0;
solve (1);
cout << ans << endl;
}
int main () {
int T = 1;
cin >> T;
while (T--) mian ();
return 0;
}
【推荐】国内首个AI IDE,深度理解中文开发场景,立即下载体验Trae
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 分享一个免费、快速、无限量使用的满血 DeepSeek R1 模型,支持深度思考和联网搜索!
· 基于 Docker 搭建 FRP 内网穿透开源项目(很简单哒)
· ollama系列01:轻松3步本地部署deepseek,普通电脑可用
· 25岁的心里话
· 按钮权限的设计及实现