HDU 6201 树形dp

题意：

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6201

给一棵n个节点的树，每个节点和边都有权值，求一条价值最大的路径。价值为该条路径终点的价值 - 起点的价值 - 路径上边的权重和。

思路：

树形dp，分别考虑祖先与后代以及兄弟之间的关系即可。细节比较多，详见代码。

代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 1e5 + 10;

struct node {
int v, w;
};

vector <node> tree[MAXN];
LL a[MAXN], dp1[MAXN], dp2[MAXN], ans;

void dfs(int u, int pa) {
dp1[u] = -INF; dp2[u] = -INF;
int t1 = -1, t2 = -1;
for (int i = 0; i < (int)tree[u].size(); i++) {
int v = tree[u][i].v, w = tree[u][i].w;
if (v == pa) continue;
dfs(v, u);
if (dp1[u] < dp1[v] - w) {
t1 = i;
dp1[u] = dp1[v] - w;
}
if (dp2[u] < dp2[v] - w) {
t2 = i;
dp2[u] = dp2[v] - w;
}
}
if (t1 != -1) ans = max(ans, dp1[u] + a[u]);
if (t2 != -1) ans = max(ans, dp2[u] - a[u]);
for (int i = 0; i < (int)tree[u].size(); i++) {
int v = tree[u][i].v, w = tree[u][i].w;
if (v == pa) continue;
if (i != t1 && t1 != -1) ans = max(ans, dp1[u] + dp2[v] - w);
if (i != t2 && t2 != -1) ans = max(ans, dp2[u] + dp1[v] - w);
}
dp1[u] = max(dp1[u], -a[u]);
dp2[u] = max(dp2[u], a[u]);
}

int main() {
//freopen("in.txt", "r", stdin);
int T;
scanf("%d", &T);
while (T--) {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%I64d", &a[i]);
tree[i].clear();
}
for (int i = 1; i < n; i++) {
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
tree[u].push_back((node){v, w});
tree[v].push_back((node){u, w});
}
ans = -INF;
dfs(1, 0);
printf("%I64d\n", ans);
}
return 0;
}