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Fire

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 1616		Accepted: 878

Description

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.

Input

The first line of input contains a single integer T representing the number of test cases. The following T blocks each represents a test case.

The first line of each block contains an integer N (1 < N <= 1000). The second line contains N numbers separated by one or more blanks. The I-th number means W(I) (0 < W(I) <= 10000). The third line contains N numbers separated by one or more blanks. The I-th number means D(I) (0 <= D(I) <= 10000). The following N-1 lines each contains three integers u, v, L (1 <= u, v <= N，0 < L <= 1000), which means there is a highway between city u and v of length L.

Output

For each test case output the minimum cost on a single line.

Sample Input

5
5
1 1 1 1 1
1 1 1 1 1
1 2 1
2 3 1
3 4 1
4 5 1
5
1 1 1 1 1
2 1 1 1 2
1 2 1
2 3 1
3 4 1
4 5 1
5
1 1 3 1 1
2 1 1 1 2
1 2 1
2 3 1
3 4 1
4 5 1
4
2 1 1 1
3 4 3 2
1 2 3
1 3 3
1 4 2
4
4 1 1 1
3 4 3 2
1 2 3
1 3 3
1 4 2

Sample Output

2
1
2
2
3

Source

POJ Monthly,Lou Tiancheng 題目鏈接：http://poj.org/problem?id=2152 題意一個國家有n個城市，城市之間的道路是樹狀結構。現在要修一些消防站，在城市建設消防站花費為w[i]，每個城市必須離消防站的距離小於d[i]。求建立完善的消防站需要至少花費多少。 思路：樹形dp。dp[u][v]表示u城市依賴v城市建設的消防站。dp[u][v]+=dp[son][v]-w[v]，其中son表示u的兒子。ans[u]表示u為根時，最優答案。 代碼：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<stack>
#include<set>
#include<bitset>
using namespace std;
#define PI acos(-1.0)
#define eps 1e-8
typedef long long ll;
typedef pair<int,int > P;
const int N=1e3+100,M=2e6+100;
const int inf=0x3f3f3f3f;
const ll INF=1e13+7,mod=1e9+7;
struct edge
{
    int from,to;
    int w;
    int next;
};
edge es[M];
int cnt,head[N];
int n;
int w[N],d[N];
int dist[N][N];
int dp[N][N],ans[N];
void init()
{
    cnt=0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w)
{
    cnt++;
    es[cnt].from=u,es[cnt].to=v;
    es[cnt].w=w;
    es[cnt].next=head[u];
    head[u]=cnt;
}
void bfs(int s)
{
    queue<int>q;
    dist[s][s]=0;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=head[u]; i!=-1; i=es[i].next)
        {
            edge e=es[i];
            if(dist[s][e.to]>dist[s][u]+e.w)
            {
                dist[s][e.to]=dist[s][u]+e.w;
                q.push(e.to);
            }
        }
    }
}
void dfs(int u,int fa)
{
    for(int i=head[u]; i!=-1; i=es[i].next)
    {
        int v=es[i].to;
        if(v==fa) continue;
        dfs(v,u);
    }
    for(int v=1; v<=n; v++)
    {
        if(dist[u][v]>d[u]) continue;
        dp[u][v]=w[v];
        for(int i=head[u]; i!=-1; i=es[i].next)
        {
            edge e=es[i];
            if(e.to==fa) continue;
            dp[u][v]+=min(dp[e.to][v]-w[v],ans[e.to]);
        }
        ans[u]=min(ans[u],dp[u][v]);
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        init();
        scanf("%d",&n);
        for(int i=1; i<=n; i++) scanf("%d",&w[i]);
        for(int i=1; i<=n; i++) scanf("%d",&d[i]);
        for(int i=1; i<n; i++)
        {
            int u,v,l;
            scanf("%d%d%d",&u,&v,&l);
            addedge(u,v,l),addedge(v,u,l);
        }
        memset(dist,inf,sizeof(dist));
        memset(dp,inf,sizeof(dp));
        memset(ans,inf,sizeof(ans));
        for(int i=1; i<=n; i++) bfs(i);
        dfs(1,0);
        printf("%d\n",ans[1]);
    }
    return 0;
}

樹形dp

POJ 2152.Fire 樹形dp