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？
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。
#include <bits/stdc++.h> #define INF 1000000007 #define FI first #define SE second #define PB emplace_back #define VI vector<int> using namespace std; typedef long long LL; typedef pair <int, int> 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; }
