题目连接：
http://acm.hdu.edu.cn/showproblem.php?pid=5834
Description
Bi Luo is a magic boy, he also has a migic tree, the tree has N nodes , in each node , there is a treasure, it's value is V[i], and for each edge, there is a cost C[i], which means every time you pass the edge i , you need to pay C[i].
You may attention that every V[i] can be taken only once, but for some C[i] , you may cost severial times.
Now, Bi Luo define ans[i] as the most value can Bi Luo gets if Bi Luo starts at node i.
Bi Luo is also an excited boy, now he wants to know every ans[i], can you help him?
Input
First line is a positive integer T(T≤104) , represents there are T test cases.
Four each test：
The first line contain an integer N(N≤105).
The next line contains N integers V[i], which means the treasure’s value of node i(1≤V[i]≤104).
Output
For the i-th test case , first output Case #i: in a single line , then output N lines , for the i-th line , output ans[i] in a single line.
Sample Input
1
5
4 1 7 7 7
1 2 6
1 3 1
2 4 8
3 5 2
Sample Output
Case #1:
15
10
14
9
15
Hint
题意
给你一棵树，边有边权，每经过边一次，就得支付过路费c[i]，点上面有宝藏，每个点只能拿一次。
问从每个点出发，能够拿到的最大值是多少？
题解：
显然的树形DP
dfs两次，第一次dfs维护从这个点的子树出去，再回来能够取得的最大值是多少，不回来的最大值是多少。
然后第二次dfs，再加上从fa那儿转移过来的值就好了，同样维护回来，和不回来。
显然答案就是max(从fa回来+从子树不回来,从fa不回来+从子树回来）
树形DP维护一下这些玩意儿就好了。
代码
// writted by dnvtmf
#include
#define INF 1000000007
#define FI first
#define SE second
#define PB emplace_back
#define VI vector
using namespace std;
typedef long long LL;
typedef pair P;
const int NUM = 100010;
struct edge {int next, to, cost;} e[NUM * 2];
int head[NUM], tot;
void gInit() {memset(head, -1, sizeof(head)); tot = 0;}
void add_edge(int u, int v, int c)
{
e[tot] = (edge) {head[u], v, c};
head[u] = tot++;
}
int n, V[NUM], w1[NUM], w2[NUM], id[NUM], ans[NUM];
//w1是从该点出发最后回到该点的最大值
//w2是从该点出发有一次不回来的最大值
//id是不回来的路的子树根节点
void dfs1(int u, int fa)
{
w1[u] = V[u];
w2[u] = V[u];
id[u] = -1;
for(int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to;
if(v == fa) continue;
dfs1(v, u);
int tmp = max(0, w1[v] - e[i].cost - e[i].cost);
int tmmp = w1[u] + max(0, w2[v] - e[i].cost);
w2[u] += tmp;
if(w2[u] < tmmp) {
w2[u] = tmmp;
id[u] = v;
}
w1[u] += tmp;
}
}
//sum1同w1,sum2同w2
void dfs2(int u, int fa, int sum1, int sum2)
{
ans[u] = max(w1[u] + sum2, w2[u] + sum1);
int W1 = w1[u];
int W2 = w2[u];
int Id = id[u];
W2 += sum1;
if(W2 <= W1 + sum2) {
W2 = W1 + sum2;
Id = fa;
}
W1 += sum1;
for(int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to;
if(v == fa) continue;
if(v == Id) {
int tmp1 = sum1 + V[u], tmp2 = sum2 + V[u];
for(int j = head[u]; ~j; j = e[j].next) {
int vv = e[j].to;
if(vv == fa || v == vv) continue;
int tmp = max(0, w1[vv] - e[j].cost - e[j].cost);
tmp2 = max(tmp1 + max(0, w2[vv] - e[j].cost), tmp2 + tmp);
tmp1 += tmp;
}
tmp1 = max(0, tmp1 - e[i].cost - e[i].cost);
tmp2 = max(0, tmp2 - e[i].cost);
dfs2(v, u, tmp1, tmp2);
}
else {
int tmp = max(0, w1[v] - e[i].cost - e[i].cost);
int tmp1 = max(0, W1 - tmp - e[i].cost - e[i].cost);
int tmp2 = max(0, W2 - tmp - e[i].cost);
dfs2(v, u, tmp1, tmp2);
}
}
}
int main()
{
int T; scanf("%d", &T);
for(int cas = 1; cas <= T; ++cas) {
gInit();
scanf("%d", &n);
for(int i = 1; i <= n; ++i) {
scanf("%d", &V[i]);
}
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:\n", cas);
for(int i = 1; i <= n; ++i)
printf("%d\n", ans[i]);
}
return 0;
}