作者：linxiuying261 | 来源：互联网 | 2023-09-18 15:51
BalancingActTimeLimit:1000MSMemoryLimit:65536KTotalSubmissions:8048Accepted:3322Descript
Balancing Act
Time Limit: 1000MS |
|
Memory Limit: 65536K |
Total Submissions: 8048 |
|
Accepted: 3322 |
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
1 #include
2 #include
3 #include
4 #include
5 #include
6 #include
7 #include
8 using namespace std;
9
10 vector<int>Q[20002];
11 int num[20002];
12 int dp[20002][2];
13 //DP[I][0] 统计子节点的最大个数
14 //DP[I][1] 统计 自身及子节点 的总个数
15 bool use[20002];
16 void add(int x,int y)
17 {
18 Q[x].push_back(y);
19 num[x]++;
20 }
21 int Max(int x,int y)
22 {
23 return x>y? x:y;
24 }
25 void dfs(int k)
26 {
27 int i,t,cur=0;
28 use[k]=true;
29 dp[k][1]=1;
30 for(i=0;i)
31 {
32 t=Q[k][i];
33 if(use[t]==true)continue;
34 dfs(t);
35
36 if(cur1])
37 cur=dp[t][1];
38 dp[k][1]=dp[k][1]+dp[t][1];
39 }
40 dp[k][0]=cur;
41 }
42 int main()
43 {
44 int T;
45 int i,n,x,y,cur,hxl,tom;
46 scanf("%d",&T);
47 while(T--)
48 {
49 scanf("%d",&n);
50 for(i=0;i<=n;i++)
51 {
52 num[i]=0;
53 Q[i].clear();
54 }
55 memset(dp,0,sizeof(dp));
56 memset(use,false,sizeof(use));
57
58 for(i=1;i)
59 {
60 scanf("%d%d",&x,&y);
61 add(x,y);
62 add(y,x);
63 }
64 dfs(1);
65 hxl=20005;
66 for(i=1;i<=n;i++)
67 {
68 cur=Max(dp[i][0],n-dp[i][1]);
69 if(cur<hxl)
70 {
71 hxl=cur;
72 tom=i;
73 }
74 }
75 printf("%d %d\n",tom,hxl);
76 }
77 return 0;
78 }
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