2022 年杭电多校第七场 1007 Weighted Beautiful Tree

Posted 2023-02-03 TURNINING

思路：考虑dp。$dp[u][0/1]$表示$u$的点权作为父边的$min/max$。考虑转移，我们需要枚举当前节点修改后的权值，显然这个权值只会是与$u$相连的所有边的边权和修改前$u$的点权中的一个。
我们只需要把子树按照边权排序，然后依次枚举点权，当与该点相连的子树的边权等于点权时需要特判取：$min(dp[v][1],dp[v[[0])$, 不相等的部分按照状态转移即可，可以通过维护两个指针实现。实现的细节看具体代码。

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<ll, int> P;

const int MAXN = 1e5 + 10;
const ll INF = 1e17;

struct edge int to, we;;

vector<edge> g[MAXN];
int n;
int c[MAXN], wn[MAXN];

ll dp[MAXN][2], pre[MAXN], sum[MAXN];

bool cmp(int x, int y)  return pre[x] < pre[y]; 

ll cal(int u, int val) 
    if(val <= wn[u]) 
        return 1ll * c[u] * (wn[u] - val);
    else 
        return 1ll * c[u] * (val - wn[u]);


void dfs(int u, int fa) 
    dp[u][1] = dp[u][0] = INF;
    vector<int> vec, val;
    for(auto e : g[u]) 
        int v = e.to;
        int we = e.we;
        if(v == fa) continue;
        pre[v] = we;
        vec.push_back(v); val.push_back(we);
        dfs(v, u);
    
    val.push_back(wn[u]);
    val.push_back(pre[u]);
    
    sort(vec.begin(), vec.end(), cmp);
    sort(val.begin(), val.end());
    int l = 0, r = 0;
    ll p = 0, s = 0;
    for(int i = 0; i < vec.size(); i++) 
        sum[i+1] = sum[i] + min(dp[vec[i]][1], dp[vec[i]][0]);
        s += dp[vec[i]][1];
     
    for(int i = 0; i < val.size(); i++) 
        while(l < vec.size() && pre[vec[l]] < val[i]) 
            p += dp[vec[l]][0];
            l++;
        
        while(r < vec.size() && pre[vec[r]] <= val[i]) 
            s -= dp[vec[r]][1];
            r++;
           
        ll w = cal(u, val[i]);
        ll c = p + s + sum[r] - sum[l] + w;
        if(val[i] <= pre[u]) dp[u][0] = min(dp[u][0], c);
        if(val[i] >= pre[u]) dp[u][1] = min(dp[u][1], c);
    


void solve() 
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) 
        g[i].clear();
    
    for(int i = 1; i <= n; i++) 
        scanf("%d", &c[i]);
    
    for(int i = 1; i <= n; i++) 
        scanf("%d", &wn[i]);
    
    for(int i = 1; i < n; i++) 
        int u, v, we; scanf("%d %d %d", &u, &v, &we);
        g[u].push_back(v, we);
        g[v].push_back(u, we);
    
    pre[1] = 0;
    dfs(1, 0);
    printf("%lld\\n", min(dp[1][0], dp[1][1]));


int main() 
    int t = 1; scanf("%d", &t);
    while(t--) 
        solve();
    
    return 0;