POJ 2152 Fire (树形DP)

题意：给定一棵树，要建立一些消防站，并且每个结点到最近一个的消防站的距离不能超过limit i，在每个结点建立消防站要花一定的费用cost i，求最少的花费是多少。

析：想了很久，确实是没想出来怎么做，dp[i][j] 表示 i 结点依赖 j 结点的最小花费，然后ans[i] 表示 以 i 为根结点的树，最少花费。在转移时，就是直接枚举 i 结点，所以依赖的 j 结点，然后ans 是取最小值，dp[i][j] += min{ dp[k][j] - cost[j], ans[k] }。其中 k 是 i 的子结点，因为 i 结点依赖 j ，其子树也是可以 j 结点的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 50;
const int mod = 1000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r > 0 && r <= n && c > 0 && c <= m;
}

struct Edge{
  int to, next, val;
};
Edge edge[maxn<<1];
int head[maxn], cnt;
int cost[maxn], limit[maxn];

void addEdge(int u, int v, int c){
  edge[cnt].to = v;
  edge[cnt].val = c;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}

int dist[maxn][maxn];
int ans[maxn], dp[maxn][maxn];

void dfs_for_dist(int u, int fa, int sp){
  for(int i = head[u]; ~i; i = edge[i].next){
    int v = edge[i].to;
    if(v == fa)  continue;
    dist[sp][v] = dist[v][sp] = dist[sp][u] + edge[i].val;
    dfs_for_dist(v, u, sp);
  }
}

void dfs(int u, int fa){
  for(int i = head[u]; ~i; i = edge[i].next){
    int v = edge[i].to;
    if(v == fa)  continue;
    dfs(v, u);
  }
  for(int j = 1; j <= n; ++j){
    if(dist[u][j] > limit[u])  continue;
    dp[u][j] = cost[j];
    for(int i = head[u]; ~i; i = edge[i].next){
      int v = edge[i].to;
      if(v == fa)  continue;
      dp[u][j] += min(ans[v], dp[v][j] - cost[j]);
    }
    ans[u] = min(ans[u], dp[u][j]);
  }
}

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)  scanf("%d", cost+i);
    for(int i = 1; i <= n; ++i)  scanf("%d", limit+i);
    ms(head, -1);  cnt = 0;
    for(int i = 1; i < n; ++i){
      int u, v, c;
      scanf("%d %d %d", &u, &v, &c);
      addEdge(u, v, c);
      addEdge(v, u, c);
    }
    for(int i = 1; i <= n; ++i)  dfs_for_dist(i, -1, i);
    ms(ans, INF);  ms(dp, INF);
    dfs(1, -1);
    printf("%d\n", ans[1]);
  }
  return 0;
}