HDU 2196 Computer(树状DP)

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给定一课树，输出每个节点的到其他节点的最远距离

dp_down[i]表示从i出发到达其子节点的最大距离

dp_up[i]表示从i出发到其父节点的最大距离

我们先更新完dp_down

然后对于每个节点我们找到最大值最小值，同时把之前已经找到最大值算进入，

然后更新dp_up

最后每个节点的最远距离为 max(dp_up,dp_down)

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <string.h>#include <algorithm>#include <cmath>#include <stack>#include <queue>#include <set>#include <vector>#include <bitset>#include <map>using namespace std;typedef long long ll;typedef pair<ll, ll> pii;const ll mod = 1000000007;const ll INF = 10000000000000000LL;typedef unsigned long long ull;const int MAXN = 100000 + 10;struct Edge{int to,next;ll cost;}edge[MAXN];int head[MAXN], tot;ll dp_down[MAXN];//从u出发到其子节点的最远距离ll dp_up[MAXN]; //从u出发到其父节点的最远距离void addedge(int u,int v,ll cost){edge[tot].to = v;edge[tot].next = head[u];edge[tot].cost = cost;head[u] = tot++;}void init(){tot = 0;memset(head, -1, sizeof head);memset(dp_down, 0, sizeof dp_down);memset(dp_up, 0, sizeof dp_up);}void dfs1(int u, int fa){//从u出发到子节点的最大值for (int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].to;if (v == fa)continue;dfs1(v, u);dp_down[u] = max(dp_down[u], dp_down[v]+edge[i].cost);}}void dfs2(int u, int fa){ll mx1 = dp_up[u], mx2 = 0,tmp;//获得从u出发到根节点的最大值与次大值，注意从u出发到父节点的值也要算for (int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].to;if (v == fa)continue;tmp=dp_down[v] + edge[i].cost;if (tmp >= mx1){mx2 = mx1;mx1 = tmp;}else if (tmp>= mx2)mx2 = tmp;}for (int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].to;if (v == fa)continue;tmp = dp_down[v] + edge[i].cost;if (tmp == mx1)dp_up[v] = mx2+edge[i].cost;else dp_up[v] = mx1+edge[i].cost;//更新节点dfs2(v, u);}}int main(){int n;while (cin >> n){init();for (int i = 2; i <=n; i++){int v;ll co;scanf("%d%lld", &v, &co);addedge(i, v, co);addedge(v, i, co);}dfs1(1, 1);dfs2(1, 1);for (int i = 1; i <= n; i++)printf("%d\n", max(dp_down[i], dp_up[i]));}}

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