题意：给定一棵树，每个点有个权值，每条边有权值，每经过边都会消耗相应的权值，但是点的权值只能获得一次，问你从 i 点出发能获得的最大权值是多少。
析：树形DP，就是太麻烦了，两次dfs，维护一共6个值分别是，从 i 出发的最大值并且返回 i， 从 i 出发的最大值并且不返回，从 i 出发的次大值并且不返回，从 i 出发的最大值的子树结点并且不返回，从 i 向父结点出发的最大值并且不返回，从 i 向父结点出发的最大值并且返回。
第一次dfs就能求出前四个，第二个dfs维护后面两个。
答案就是 max(所有子节点返回该节点+ 父亲节点不返回该节点， 有个子节点不返回该节点+从父亲节点返回该节点)；
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>
#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 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 LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
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;
}
int val[maxn];
struct Edge{
int to, next, c;
};
Edge edge[maxn<<1];
int head[maxn], cnt;
int dp[maxn][4];
int ans[maxn];
/*
dp[i][0] the most val and back
dp[i][1] the most val and not back
dp[i][2] the second most val and not back
dp[i][3] the the son of the dp[i][1]
*/
void add_edge(int u, int v, int val){
edge[cnt].to = v;
edge[cnt].c = val;
edge[cnt].next = head[u];
head[u] = cnt++;
}
void dfs1(int u, int fa){
dp[u][0] = dp[u][1] = val[u];
dp[u][2] = 0;
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa) continue;
dfs1(v, u);
}
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa) continue;
int c = edge[i].c;
dp[u][0] += max(0, dp[v][0] - 2*c);
}
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa) continue;
int c = edge[i].c;
int val = dp[u][0] - max(0, dp[v][0] - 2*c) + max(0, dp[v][1] - c);
if(val > dp[u][1]){
dp[u][2] = dp[u][1];
dp[u][1] = val;
dp[u][3] = v;
}
else if(val > dp[u][2])
dp[u][2] = val;
}
}
void dfs2(int u, int fa, int fb, int fnb){
ans[u] = max(fb+dp[u][1], fnb+dp[u][0]);
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa) continue;
int c = edge[i].c;
int newfb = fb + dp[u][0] - max(0, dp[v][0] - 2*c);
int newfnb;
if(v == dp[u][3])
newfnb = max(fb + dp[u][2] - max(0, dp[v][0]-2*c), fnb + dp[u][0] - max(0, dp[v][0]-2*c));
else
newfnb = max(fb + dp[u][1] - max(0, dp[v][0]-2*c), fnb + dp[u][0] - max(0, dp[v][0]-2*c));
dfs2(v, u, max(0, newfb-2*c), max(0, newfnb-c));
}
}
int main(){
int T; cin >> T;
for(int kase = 1; kase <= T; ++kase){
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", val+i);
memset(head, -1, sizeof head);
cnt = 0;
for(int i = 1; i < n; ++i){
int u, v, c;
scanf("%d %d %d", &u, &v, &c);
add_edge(u, v, c);
add_edge(v, u, c);
}
dfs1(1, -1);
dfs2(1, -1, 0, 0);
printf("Case #%d:
", kase);
for(int i = 1; i <= n; ++i)
printf("%d
", ans[i]);
}
return 0;
}