hzau 1201 Friends（树形dp）

1201: Friends

Time Limit: 1 Sec Memory Limit: 1280 MB
Submit: 119 Solved: 25
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Description

In a country, the relationship between people can be indicated by a tree. If two people are acquainted with each other, there will be an edge between them. If a person can get in touch with another through no more than five people, we should consider these two people can become friends. Now, give you a tree of N people’s relationship. ( 1 <= N <= 1e5), you should compute the number of who can become friends of each people?

Input

In the first line, there is an integer T, indicates the number of the cases.
For each case, there is an integer N indicates the number of people in the first line.

In the next N-1 lines, there are two integers u and v, indicate the people u and the people

v are acquainted with each other directly. People labels from 1.

Output

For each case, the first line is “Case #k :”, k indicates the case number.

In the next N lines, there is an integer in each line, indicates the number of people who can become the ith person’s friends. i is from 1 to n.

Sample Input

Sample Output

Case #1:
6
7
7
7
7
7
7
6

HINT

树形dp，

ch[u][i]数组表示u节点到达其孩子的距离为i的点的个数，
　　ch[u][i]=sum(dpch[v][i-1])，

fa[u][i]表示u节点向上通过父节点到达的距离为i的点的个数，
　　fa[u][i]=fa[fa][i-1]+ch[fa][i-1]-ch[u][i-2]，
两次dfs计算出这两个值

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int MAXN = 1e5 + 5;
 5 
 6 vector<int> vt[MAXN];
 7 
 8 int fa[MAXN][10];//u节点到达其孩子的距离为i的点的个数
 9 int ch[MAXN][10];//u节点向上通过父节点到达的距离为i的点的个数
10 
11 void dfs(int u, int fa)
12 {
13     ch[u][0] = 1;
14     if (vt[u].size() == 0) {
15         return;
16     }
17     int i, j;
18     int v;
19     for (i = 0; i < vt[u].size(); ++i) {
20         v = vt[u][i];
21         if (v != fa) {
22             dfs(v, u);
23             for (j = 1; j <= 6; ++j) {
24                 ch[u][j] += ch[v][j - 1];
25             }
26         }
27     }
28 }
29 
30 void dp(int u, int fa)
31 {
32     fa[u][0] = 1;
33     int i;
34 
35     if (u != fa) {//非根节点
36         fa[u][1] = 1;
37         for (i = 2; i <= 6; ++i) {
38             fa[u][i] = fa[fa][i - 1] + ch[fa][i - 1] - ch[u][i - 2];
39         }
40     }
41     int v;
42     for (i = 0; i < vt[u].size(); ++i) {
43         v = vt[u][i];
44         if (v != fa) {
45             dp(v, u);
46         }
47     }
48 }
49 
50 int main()
51 {
52     int T;
53     int n;
54     int u, v;
55     int i, j;
56     int cas = 0;
57     int ans;
58 
59     scanf("%d", &T);
60     while (T--) {
61         scanf("%d", &n);
62         for (i = 1; i <= n; ++i) {
63             vt[i].clear();
64         }
65         memset(fa, 0, sizeof(fa[0]) * (n + 1));
66         memset(ch, 0, sizeof(ch[0]) * (n + 1));
67         for (i = 0; i < n - 1; ++i) {
68             scanf("%d%d", &u, &v);
69             vt[u].push_back(v);
70             vt[v].push_back(u);
71         }
72         dfs(1, 1);
73         dp(1, 1);
74         printf("Case #%d:
", ++cas);
75 
76 
77         for (i = 1; i <= n; ++i) {
78             ans = 0;
79             for (j = 1; j <= 6; ++j) {
80                 ans += fa[i][j] + ch[i][j];
81             }
82             printf("%d
", ans);
83         }
84     }
85 
86     return 0;
87 }