题目链接
题解
经典老题，还真暴力
\(n \le 1000\)，所以可以\(O(n^2)\)做
所以可以枚举每个点依附于哪一个点
设\(f[u]\)表示以\(u\)为根的子树的最小代价
\(g[u][v]\)表示\(u\)依附于\(v\)时以\(u\)为根的子树的最小代价
显然
\[f[u] = min\{ g[u][v] \}\]
\[g[u][v] = cost[v] + \sum\limits_{(u,to) \in edge} min(g[to][v] - cost[v],f[to]) \quad [dis(u,v) \le D(u)]\]
\(dis(u,v)\)要直接暴力\(O(n^2)\)预处理
上倍增直接\(T\)掉。。。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define cls(s) memset(s,0,sizeof(s))
using namespace std;
const int maxn = 1005,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int h[maxn],ne;
struct EDGE{int to,nxt,w;}ed[maxn << 1];
inline void build(int u,int v,int w){
ed[ne] = (EDGE){v,h[u],w}; h[u] = ne++;
ed[ne] = (EDGE){u,h[v],w}; h[v] = ne++;
}
int fa[maxn],f[maxn],g[maxn][maxn],cost[maxn],d[maxn],dis[maxn][maxn],n,rt;
void DFS(int u,int D,int F){
dis[rt][u] = D;
Redge(u) if ((to = ed[k].to) != F)
DFS(to,D + ed[k].w,u);
}
void dfs(int u){
f[u] = INF;
REP(i,n)
if (dis[u][i] <= d[u]) g[u][i] = cost[i];
else g[u][i] = INF;
Redge(u) if ((to = ed[k].to) != fa[u]){
fa[to] = u; dfs(to);
REP(i,n) if (g[u][i] != INF)
g[u][i] += min(g[to][i] - cost[i],f[to]);
}
REP(i,n) f[u] = min(f[u],g[u][i]);
}
int main(){
int T = read();
while (T--){
n = read(); ne = 2; cls(h);
REP(i,n) cost[i] = read();
REP(i,n) d[i] = read();
int a,b,w;
for (int i = 1; i < n; i++){
a = read(); b = read(); w = read();
build(a,b,w);
}
REP(i,n) rt = i,DFS(i,0,0);
dfs(1);
printf("%d\n",f[1]);
}
return 0;
}