hdu 2196 Computer 树形DP

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2196

A school bought the first computer some time ago(so this computer's id is 1). During the recent years the school bought N-1 new computers. Each new computer was connected to one of settled earlier. Managers of school are anxious about slow functioning of the net and want to know the maximum distance Si for which i-th computer needs to send signal (i.e. length of cable to the most distant computer). You need to provide this information. 

Hint: the example input is corresponding to this graph. And from the graph, you can see that the computer 4 is farthest one from 1, so S1 = 3. Computer 4 and 5 are the farthest ones from 2, so S2 = 2. Computer 5 is the farthest one from 3, so S3 = 3. we also get S4 = 4, S5 = 4.

题意描述：给出一颗树，求出每个节点到叶节点的最长距离。

算法分析：树形DP。每个节点向下dfs一遍，向上dfs一遍即可。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<vector>
 9 #define inf 0x7fffffff
10 using namespace std;
11 const int maxn=10000+10;
12 
13 int n;
14 int fmaxn[maxn],fmaxnid[maxn],smaxn[maxn],smaxnid[maxn];
15 vector<pair<int,int> >vec[maxn];
16 
17 void dfs(int u,int pre)
18 {
19     int k=vec[u].size();
20     for (int i=0 ;i<k ;i++)
21     {
22         int v=vec[u][i].first;
23         int w=vec[u][i].second;
24         if (v==pre) continue;
25         dfs(v,u);
26         if (smaxn[u]<fmaxn[v]+w)
27         {
28             smaxn[u]=fmaxn[v]+w;
29             smaxnid[u]=v;
30             if (smaxn[u]>fmaxn[u])
31             {
32                 swap(fmaxn[u],smaxn[u]);
33                 swap(fmaxnid[u],smaxnid[u]);
34             }
35         }
36     }
37     return;
38 }
39 
40 void dfs2(int u,int pre)
41 {
42     int k=vec[u].size();
43     for (int i=0 ;i<k ;i++)
44     {
45         int v=vec[u][i].first;
46         int w=vec[u][i].second;
47         if (v==pre) continue;
48         if (v==fmaxnid[u])
49         {
50             if (smaxn[u]+w>smaxn[v])
51             {
52                 smaxn[v]=smaxn[u]+w;
53                 smaxnid[v]=u;
54                 if (smaxn[v]>fmaxn[v])
55                 {
56                     swap(smaxn[v],fmaxn[v]);
57                     swap(smaxnid[v],fmaxnid[v]);
58                 }
59             }
60         }
61         else if (fmaxn[u]+w>smaxn[v])
62         {
63             smaxn[v]=fmaxn[u]+w;
64             smaxnid[v]=u;
65             if (smaxn[v]>fmaxn[v])
66             {
67                 swap(smaxn[v],fmaxn[v]);
68                 swap(smaxnid[v],fmaxnid[v]);
69             }
70         }
71         dfs2(v,u);
72     }
73     return ;
74 }
75 
76 int main()
77 {
78     while (scanf("%d",&n)!=EOF)
79     {
80         for (int i=1 ;i<=n ;i++) vec[i].clear();
81         memset(fmaxn,0,sizeof(fmaxn));
82         memset(fmaxnid,0,sizeof(fmaxnid));
83         memset(smaxn,0,sizeof(smaxn));
84         memset(smaxnid,0,sizeof(smaxnid));
85         int a,b;
86         for (int i=2 ;i<=n ;i++)
87         {
88             scanf("%d%d",&a,&b);
89             vec[i].push_back(make_pair(a,b));
90             vec[a].push_back(make_pair(i,b));
91         }
92         dfs(1,1);
93         dfs2(1,1);
94         for (int i=1 ;i<=n ;i++) printf("%d\n",fmaxn[i]);
95     }
96     return 0;
97 }

 

转载于:https://www.cnblogs.com/huangxf/p/4370159.html