POJ1655 Balancing Art

Balancing Act

Description

Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or more trees. Define the balance of a node to be the size of the largest tree in the forest T created by deleting that node from T.
For example, consider the tree:

Deleting node 4 yields two trees whose member nodes are {5} and
{1,2,3,6,7}. The larger of these two trees has five nodes, thus the
balance of node 4 is five. Deleting node 1 yields a forest of three
trees of equal size: {2,6}, {3,7}, and {4,5}. Each of these trees has
two nodes, so the balance of node 1 is two.

For each input tree, calculate the node that has the minimum
balance. If multiple nodes have equal balance, output the one with the
lowest number.

Input

The
first line of input contains a single integer t (1 <= t <= 20),
the number of test cases. The first line of each test case contains an
integer N (1 <= N <= 20,000), the number of congruence. The next
N-1 lines each contains two space-separated node numbers that are the
endpoints of an edge in the tree. No edge will be listed twice, and all
edges will be listed.

Output

For
each test case, print a line containing two integers, the number of the
node with minimum balance and the balance of that node.

Sample Input

1
7
2 6
1 2
1 4
4 5
3 7
3 1

Sample Output

1 2

Source

POJ Monthly--2004.05.15 IOI 2003 sample task

【题解】

此题用所谓“DP”？求重心即可，更新时时刻注意更新最小节点

一直不理解为啥求重心也叫DP啊！

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5
 6 inline void read(int &x)
 7 {
 8     x = 0;char ch = getchar();char c = ch;
 9     while(ch > ‘9‘ || ch < ‘0‘)c = ch, ch = getchar();
10     while(ch <= ‘9‘ && ch >= ‘0‘)x = x * 10 + ch - ‘0‘, ch = getchar();
11     if(c == ‘-‘)x = -x;
12 }
13 inline int max(int a, int b){return a > b ? a : b;}
14 inline int min(int a, int b){return a < b ? a : b;}
15
16 const int INF = 0x3f3f3f3f;
17 const int MAXN = 200000 + 10;
18
19 struct Edge
20 {
21     int u,v,next;
22 }edge[MAXN << 1];
23 int t,n,head[MAXN],cnt,b[MAXN],dp[MAXN],ma,g;
24 inline void insert(int a, int b){edge[++cnt] = Edge{a,b,head[a]};head[a] = cnt;}
25
26 void dfs(int u)
27 {
28     int pos, tmp = -1;
29     dp[u] = 1;
30     for(pos = head[u];pos;pos = edge[pos].next)
31     {
32         int v = edge[pos].v;
33         if(!b[v])
34         {
35             b[v] = true;
36             dfs(v);
37             dp[u] += dp[v];
38             tmp = max(tmp, dp[v]);
39         }
40     }
41     tmp = max(tmp, n - dp[u]);
42     if(tmp < ma)
43         ma = tmp,g = u;
44     if(tmp == ma)
45         g = min(g, u);
46 }
47
48 int main()
49 {
50     read(t);
51     register int i,tmp1,tmp2;
52     for(;t;--t)
53     {
54         ma = INF,g = INF;
55         memset(edge, 0, sizeof(edge));
56         memset(head, 0, sizeof(head));
57         cnt = 0;memset(dp, 0, sizeof(dp));
58         memset(b, 0, sizeof(b));
59         read(n);
60         for(i = 1;i < n;++ i)
61         {
62             read(tmp1);read(tmp2);
63             insert(tmp1, tmp2);insert(tmp2, tmp1);
64         }
65         b[1] = true;
66         dfs(1);
67         printf("%d %d\n", g, ma);
68     }
69     return 0;
70 }