POJ1655 Balancing Art

Posted 2020-09-26 嘒彼小星

Balancing Act

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13865		Accepted: 5880

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 }

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