Magic boy Bi Luo with his excited tree
Problem 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).
For the next N−1 lines, each contains three integers u,v,c , which means node u and node v are connected by an edge, it‘s cost is c(1≤c≤104).
You can assume that the sum of N will not exceed 106.
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
Author
UESTC
Source
题意：
给出一棵树，每个点当经过它时有一个获益，每个点的获益只能计算一次，每条边当经过它都有一个花费，需要重复计算。
询问从每个点出发，随意走的最大收益。
题解：
树形dp。
比赛时没有A掉。。。队友没有采取我的思路写，按照他们推的公式写了。。。
（其实是当时我在写其他题，我写完他们直接上了。。。实际上不用那么急，
（当时比赛结束还有1.5h，好好整理思路写说不定就过了
其实我觉得我的思路是比较简单明了的。。
今天写了一下，从整理题目思路在内到AC，只花了1h，一发就AC了。。。
设f[i]代表每个点出去随便走不一定回来的最大收益和次大收益（一对数），
个g[i]代表每个点出去随便走一定要回来的最大收益。
按照bfs序扫两遍。
第一遍，按照bfs序倒着扫：
统计每个点只能往下走的 f 和 g。
这部分是非常简单的dp，方程
g[i] = sigma(max(0, g[child] - 2 * costOfEdge))
f[i] = max(g[i],
g[i] - max(0, g[child] - 2 * costOfEdge) + f[child] + costOfEdge)
计算f时先将孩子对g[i]的贡献减去加上本次贡献就行了。
第二遍，按照bfs序正着扫：
统计每个点的f和g，加入父亲的影响。
也就是说此时的f和g的意义变为不仅仅可以往下走还可以往父亲边走，就是随便走。
统计时类似，先减去自己对父亲的影响，再利用父亲的f和g更新自己。
转移方程这里写的不方便，具体可以查看代码。
最后每个点的f值就是答案。
因为f值不会比g值小，这点可以从定义或者方程看出。
1 const int N = 100010;
2 int head[N], son[N * 2], val[N * 2], nex[N * 2], tot;
3 int w[N], n;
4 struct FirstAndSecondMax {
5 pair<int, int> first, second;
6
7 inline void init(int val, int id) {
8 first = second = make_pair(val, id);
9 }
10
11 inline void addAll(int w) {
12 first.first += w, second.first += w;
13 }
14
15 inline void updata(int val, int id) {
16 pair<int, int> now = make_pair(val, id);
17 if(first < now) swap(first, now);
18 if(second < now) swap(second, now);
19 }
20
21 inline int getMax(int except) {
22 return first.second == except ? second.first : first.first;
23 }
24 } f[N];
25 int g[N];
26 // f -> maximum of paths that need not go back
27 // g -> maximum of paths that need to go back
28 #define cg(origin, cost) (max(0, origin - 2 * cost))
29 #define cf(origin, cost) (max(0, origin - cost))
30 // cg -> contribution to g
31 // cf -> contribution to f
32 int ans[N];
33
34 inline void init() {
35 for(int i = 0; i < n; ++i) f[i].init(w[i], i);
36 for(int i = 0; i < n; ++i) head[i] = -1;
37 tot = 0;
38 }
39
40 inline void addEdge(int u, int v, int w) {
41 son[tot] = v, val[tot] = w, nex[tot] = head[u];
42 head[u] = tot++;
43 }
44
45 int que[N], fa[N], valfa[N];
46 inline void bfs(int st) {
47 int len = 0;
48 fa[st] = -1, valfa[st] = INF, que[len++] = st;
49 for(int idx = 0; idx < len; ++idx) {
50 int u = que[idx];
51 for(int tab = head[u], v; tab != -1; tab = nex[tab])
52 if((v = son[tab]) != fa[u])
53 fa[v] = u, valfa[v] = val[tab], que[len++] = v;
54 }
55 }
56
57 inline void solve() {
58 bfs(1);
59
60 for(int idx = n - 1; idx >= 0; --idx) {
61 int u = que[idx];
62 g[u] = w[u];
63 for(int tab = head[u], v; tab != -1; tab = nex[tab])
64 if((v = son[tab]) != fa[u])
65 g[u] += cg(g[v], val[tab]);
66 f[u].init(g[u], u);
67 for(int tab = head[u], v; tab != -1; tab = nex[tab])
68 if((v = son[tab]) != fa[u])
69 f[u].updata(g[u] - cg(g[v], val[tab]) +
70 f[v].getMax(u) - val[tab], v);
71 }
72
73 for(int idx = 0; idx < n; ++idx) {
74 int u = que[idx];
75 if(fa[u] != -1) {
76 int faContributeIt = g[fa[u]] - cg(g[u], valfa[u]);
77 int lastg = g[u];
78 g[u] += cg(faContributeIt, valfa[u]);
79 f[u].addAll(cg(faContributeIt, valfa[u]));
80 f[u].updata(f[fa[u]].getMax(u) - cg(lastg, valfa[u])
81 - valfa[u] + lastg, fa[u]);
82 }
83 ans[u] = f[u].getMax(-1);
84 }
85
86 for(int i = 0; i < n; ++i) printf("%d\n", ans[i]);
87 }
88
89 int main() {
90 int testCase;
91 scanf("%d", &testCase);
92 for(int testIndex = 1; testIndex <= testCase; ++testIndex) {
93 printf("Case #%d:\n", testIndex);
94 scanf("%d", &n);
95 for(int i = 0; i < n; ++i) scanf("%d", &w[i]);
96 init();
97 for(int i = 1, u, v, w; i < n; ++i) {
98 scanf("%d%d%d", &u, &v, &w);
99 --u, --v;
100 addEdge(u, v, w), addEdge(v, u, w);
101 }
102 solve();
103 }
104 return 0;
105 }