Computer

**Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6378 Accepted Submission(s): 3211
**

Problem Description

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.

Input

Input file contains multiple test cases.In each case there is natural number N (N<=10000) in the first line, followed by (N-1) lines with descriptions of computers. i-th line contains two natural numbers - number of computer, to which i-th computer is connected and length of cable used for connection. Total length of cable does not exceed 10^9. Numbers in lines of input are separated by a space.

Output

For each case output N lines. i-th line must contain number Si for i-th computer (1<=i<=N).

Sample Input

1 1

2 1

3 1

1 1

Sample Output

Author

scnu

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题意：给n个点，1为根节点，给你的是棵树，然后找到各个节点离树上其他节点最远的距离;

思路：树形dp；

因为这是一颗树，除了根节点外其他每个节点都又且一个父亲节点，首先dfs更新每个节点到其子节点下端的最长距离和次长距离，然后最长距离还可能从父亲节点那更新。那么再dfs从父亲节点更新子节点的最长距离，这个时候，父亲节点的最长可能是通过当前要更新的子节点，那么就用次长路去更新。复杂度O（n）；

1 #include
2 #include
3 #include
4 #include
5 #include
6 #include
7 #include
8 #include
9 #include
10 #include
11 #include
12 #include
13 using namespace std;
14 typedef struct acc
15 {
16 int id;
17 int val;
18 } ak;
19 vectorvec[100005];
20 bool vis[100005];
21 typedef struct node
22 {
23 int id;
24 int mcost;
25 int scost;
26 int mid;
27 int sid;
28 node()
29 {
30 mcost = 0;
31 scost = 0;
32 mid = -1;
33 sid = -1;
34 }
35 } ss;
36 ss dp[100005];
37 int father[1000005];
38 void dfs1(int n);
39 void dfs2(int n);
40 int main(void)
41 {
42 int n,m;
43 while(scanf("%d",&n)!=EOF)
44 {
45 memset(father,-1,sizeof(father));
46 int i,j;
47 memset(dp,0,sizeof(dp));
48 for(i = 0; i < 100005; i++) 49 vec[i].clear(); 50 for(i = 2; i <= n; i++) 51 { 52 int id,val; 53 scanf("%d %d",&id,&val); 54 ak a; 55 a.val = val; 56 a.id = i; 57 father[i] = id; 58 vec[id].push_back(a); 59 } 60 dfs1(1); 61 dfs2(1); 62 for(i = 1; i <= n; i++) 63 { 64 printf("%d\n",dp[i].mcost); 65 } 66 } 67 return 0; 68 } 69 void dfs1(int n) 70 { 71 for(int i = 0; i < vec[n].size(); i++) 72 { 73 ak d = vec[n][i]; 74 { 75 dfs1(d.id); 76 if(dp[d.id].mcost+d.val > dp[n].mcost)
77 {
78 dp[n].scost = dp[n].mcost;
79 dp[n].sid = dp[n].mid;
80 dp[n].mcost = dp[d.id].mcost+d.val;
81 dp[n].mid = d.id;
82 }
83 else if(dp[d.id].mcost+d.val > dp[n].scost)
84 {
85 dp[n].scost = dp[d.id].mcost+d.val;
86 dp[n].sid = d.id;
87 }
88 }
89 }
90 }
91 void dfs2(int n)
92 {
93 for(int i = 0; i < vec[n].size(); i++) 94 { 95 ak d = vec[n][i]; 96 if(dp[n].mid!=d.id) 97 { 98 if(dp[n].mcost + d.val > dp[d.id].mcost)
99 {
100 dp[d.id].scost = dp[d.id].mcost;
101 dp[d.id].sid = dp[d.id].mid;
102 dp[d.id].mcost = dp[n].mcost + d.val;
103 dp[d.id].mid = n;
104 }
105 else if(dp[n].mcost + d.val > dp[d.id].scost)
106 {
107 dp[d.id].scost = dp[n].mcost + d.val;
108 dp[d.id].sid = n;
109 }
110 }
111 else
112 {
113 if(dp[n].scost + d.val > dp[d.id].mcost)
114 {
115 dp[d.id].scost = dp[d.id].mcost;
116 dp[d.id].sid = dp[d.id].mid;
117 dp[d.id].mcost = dp[n].scost + d.val;
118 dp[d.id].mid = n;
119 }
120 else if(dp[n].scost + d.val > dp[d.id].scost)
121 {
122 dp[d.id].scost = dp[n].scost + d.val;
123 dp[d.id].sid = n;
124 }
125 }
126 dfs2(d.id);
127 }
128 }

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