首页 > 代码库 > poj 1655 Balancing Act
poj 1655 Balancing Act
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 172 61 21 44 53 73 1 Sample Output 1 2 Source POJ Monthly--2004.05.15 IOI 2003 sample ta |
树的重心
#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int maxn = 20003;int m;struct node{ int v,next;}edge[100000];int head[maxn],num;void add_edge(int x ,int y){ edge[++num].v=y;edge[num].next=head[x];head[x]=num;}bool vis[maxn];int son[maxn];int siz=0x7fffffff,ans;void dfs(int x){ vis[x]=1; son[x]=0; int tmp=0; for(int i=head[x];i;i=edge[i].next) { int v=edge[i].v; if(vis[v])continue; dfs(v); son[x]+=(son[v]+1); tmp=max(son[v]+1,tmp); } tmp=max(tmp,m-son[x]-1); if(tmp<siz || tmp==siz && ans>x) { ans=x; siz=tmp; } }int main(){ int t; scanf("%d",&t); while(t--) { memset(vis,0,sizeof vis);num=0;siz=0x7fffffff;ans=0; memset(head,0,sizeof(head)); scanf("%d",&m); int a,b; for(int i=1;i<m;i++) { scanf("%d%d",&a,&b); add_edge(a,b);add_edge(b,a); } dfs(1); printf("%d %d\n",ans,siz); } return 0;}
poj 1655 Balancing Act
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。