https://cn.vjudge.net/contest/264707#problem/D
Country Z has N cities, which are numbered from 1 to N. Cities are connected by highways, and there is exact one path between two different cities. Recently country Z often caught fire, so the government decided to build some firehouses in some cities. Build a firehouse in city K cost W(K). W for different cities may be different. If there is not firehouse in city K, the distance between it and the nearest city which has a firehouse, can’t be more than D(K). D for different cities also may be different. To save money, the government wants you to calculate the minimum cost to build firehouses.

題意為一個n個城市的國家，編號為1-n，每個城市都有著火的風險，需要在給出的距離記憶體在一個滅火站，滅火站需要建在城市裡面且不同城市花費不同，求滿足條件的最少花費
剛開始思路是用二位存第i個城市是否修建滅火站，發現寫出來無法將距離之內的城市建站對該城市的狀態表達出來
正解是dp[i][j]表示第i個城市對其管轄的滅火站建在第j個城市，由於需要計算在給出範圍的其他城市的建站情況，所以對每個城市需要計算一次到其他城市的距離，用bst陣列記錄每個點已經算出來的最優解

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <string>
using namespace std;

const int maxn = 1e3 + 4;
const int inf = 0x3f3f3f3f;
vector< pair<int, int> > e[maxn];
vector<int> w(maxn), d(maxn), bst(maxn), dis(maxn);
int dp[maxn][maxn];
int t, n;
void Dis(int u) {
    for(int i = 0; i < e[u].size(); i++) {
        int v = e[u][i].first, l = e[u][i].second;
        if(dis[v] != -1) continue;
        dis[v] = dis[u] + l;
        Dis(v);
    }
}
void dfs(int u, int pre) {
    for(int i = 0; i < e[u].size(); i++) {
        int v = e[u][i].first, l = e[u][i].second;
        if(v == pre) continue;
        dfs(v, u);
    }
    dis.assign(n + 1, -1);
    dis[u] = 0;
    bst[u] = inf;
    Dis(u);
    for(int i = 1; i <= n; i++) {
        if(dis[i] > d[u]) dp[u][i] = inf;
        else {
            dp[u][i] = w[i];
            for(int j = 0; j < e[u].size(); j++) {
                int v = e[u][j].first, l = e[u][j].second;
                if(v == pre) continue;
                dp[u][i] += min(bst[v], dp[v][i] - w[i]);
            }
            bst[u] = min(bst[u], dp[u][i]);
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> t;
    while(t--) {
        cin >> n;
        for(int i = 1; i <= n; i++) e[i].clear();
        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; i++) {
            int u, v, l;
            cin  >> u >> v >> l;
            e[u].push_back(make_pair(v, l));
            e[v].push_back(make_pair(u, l));
        }
        dfs(1, 0);
        cout << bst[1] << endl;
    }
}