[poj 2152] fire (树形dp)

Time Limit: 2000MS Memory Limit: 65536K
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

很好的树形dp~
f[u][i] 表示u点及其子树的所有节点都被保护且i来保护u的最优值
ans[i] 表示i及其子树所有节点都被保护的最优值
那么显然ans[i]=min(ans[i],f[i][j]) (1<=j<=n)
思路：
显然是从叶节点不断向上更新至根 则最后结果为ans[1]
要求一个节点(u)的ans值那么就要求出所有f[u][i] (1<=i<=n) 取min ，
那么对于其中一种情况 要求f[u][i] 时（u、i值固定），需要遍历所有u的子节点并把所有子节点影响u的最小花费（可能是原来它对应的ans，也可能是如果选i点那么i点可以一举两得使一个cost[i]保护多个点并且优于取ans）加和 再加上cost[i]
出错（我第一次写时犯的错qwq 各种脑残233）：
在dfs中遍历u所有节点时把’j’写成了‘i’（代码中有标注），还查了将近半个小时
ヽ(｀Д´)ﾉ︵ ┻━┻ ┻━┻

code:

//By Menteur_Hxy
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;

const int MAX=1010;
const int INF=0x3f3f3f3f;
int n,cnt;
int cost[MAX],d[MAX],head[MAX],dis[MAX][MAX],dp[MAX][MAX],ans[MAX];

int rd() {
    int x=0;
    char c=' ';
    while(c==' ' || c=='
') c=getchar();
    while(c<='9'&&c>='0') {
        x=x*10+c-'0';
        c=getchar();
    }
    return x;
}

struct edges{
    int to,next,dis;
    void add(int a,int b,int c) {
        to=b,next=head[a],dis=c,head[a]=cnt;
    }
}edge[MAX<<1];

queue <int> q;

void getd(int dis[],int s) {
    dis[s]=0;
    q.push(s);
    while(!q.empty()) {
        int u=q.front(); q.pop();
        for(int i=head[u];i;i=edge[i].next) {
            int v=edge[i].to;
            if(!dis[v] && v!=s) {
                dis[v]=dis[u]+edge[i].dis;
                q.push(v);
            }
        }
    }
}

void dfs(int u,int fa) {
    for(int i=head[u];i;i=edge[i].next) 
        if(edge[i].to!=fa) dfs(edge[i].to,u);

    for(int i=1;i<=n;i++) {
        if(dis[u][i]<=d[u]) {
            int tmp=0;
            for(int j=head[u];j;j=edge[j].next) {
                int v=edge[j].to;//出错处233 j写成i
                if(v!=fa) tmp+=min(ans[v],dp[v][i]-cost[i]);
            }
            dp[u][i]=tmp+cost[i];
        }
        else dp[u][i]=INF;
    }
    for(int i=1;i<=n;i++) ans[u]=min(ans[u],dp[u][i]);
}

void init() {
    cnt=0;
    memset(head,0,sizeof head);
    memset(ans,0x3f,sizeof ans);
    memset(dp,0,sizeof dp);
    memset(dis,0,sizeof dis);
}

int main() {
    int T=rd();
    while(T--) {
        init();
        n=rd();
        for(int i=1;i<=n;i++) cost[i]=rd();
        for(int i=1;i<=n;i++) d[i]=rd();
        for(int i=1;i<n;i++) {
            int a=rd(),b=rd(),c=rd();
            edge[++cnt].add(a,b,c);
            edge[++cnt].add(b,a,c);
        }
        for(int i=1;i<=n;i++) getd(dis[i],i);
        dfs(1,-1);
        printf("%d
",ans[1]);
    }
    return 0;
}