FZU 2256 树形dp（水）

题意：

题目链接：http://acm.fzu.edu.cn/problem.php?pid=2256
中文题。

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思路：

一遍dfs，维护路径中遇到的传送门的最短时间即可。

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代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>using namespace std;typedef long long LL;const int MAXN = 1e5 + 10;const LL INF = 1e15 + 10;struct node {    LL v, w;};LL dp[MAXN], d[MAXN];vector <node> tree[MAXN];void dfs(int u, LL Mintime) {    int cnt = tree[u].size();    for (int i = 0; i < cnt; i++) {        LL v = tree[u][i].v, w = tree[u][i].w;        dp[v] = min(Mintime, dp[u] + w);        dfs(v, min(dp[v] + d[v], Mintime));    }}int main() {    //freopen("in.txt", "r", stdin);    int n;    while (scanf("%d", &n) != EOF) {        for (int i = 1; i <= n; i++) {            tree[i].clear();            scanf("%lld", &d[i]);        }        for (int i = 1; i < n; i++) {            LL p, w;            scanf("%lld%lld", &p, &w);            tree[p].push_back((node){i + 1, w});        }        dp[1] = 0;        dfs(1, d[1]);        for (int i = 1; i <= n; i++)            printf("%lld ", dp[i]);        puts("");    }    return 0;}