POJ2378(树形DP)
2016-02-27 22:26
302 查看
Tree Cutting
Description
After Farmer John realized that Bessie had installed a "tree-shaped" network among his N (1 <= N <= 10,000) barns at an incredible cost, he sued Bessie to mitigate his losses.
Bessie, feeling vindictive, decided to sabotage Farmer John's network by cutting power to one of the barns (thereby disrupting all the connections involving that barn). When Bessie does this, it breaks the network into smaller pieces, each of which retains full connectivity within itself. In order to be as disruptive as possible, Bessie wants to make sure that each of these pieces connects together no more than half the barns on FJ.
Please help Bessie determine all of the barns that would be suitable to disconnect.
Input
* Line 1: A single integer, N. The barns are numbered 1..N.
* Lines 2..N: Each line contains two integers X and Y and represents a connection between barns X and Y.
Output
* Lines 1..?: Each line contains a single integer, the number (from 1..N) of a barn whose removal splits the network into pieces each having at most half the original number of barns. Output the barns in increasing numerical order. If there are no suitable barns, the output should be a single line containing the word "NONE".
Sample Input
Sample Output
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 4156 | Accepted: 2512 |
After Farmer John realized that Bessie had installed a "tree-shaped" network among his N (1 <= N <= 10,000) barns at an incredible cost, he sued Bessie to mitigate his losses.
Bessie, feeling vindictive, decided to sabotage Farmer John's network by cutting power to one of the barns (thereby disrupting all the connections involving that barn). When Bessie does this, it breaks the network into smaller pieces, each of which retains full connectivity within itself. In order to be as disruptive as possible, Bessie wants to make sure that each of these pieces connects together no more than half the barns on FJ.
Please help Bessie determine all of the barns that would be suitable to disconnect.
Input
* Line 1: A single integer, N. The barns are numbered 1..N.
* Lines 2..N: Each line contains two integers X and Y and represents a connection between barns X and Y.
Output
* Lines 1..?: Each line contains a single integer, the number (from 1..N) of a barn whose removal splits the network into pieces each having at most half the original number of barns. Output the barns in increasing numerical order. If there are no suitable barns, the output should be a single line containing the word "NONE".
Sample Input
10 1 2 2 3 3 4 4 5 6 7 7 8 8 9 9 10 3 8
Sample Output
3 8 题意:问删除哪些结点后,剩余各个子树的大小均不超过总结点数目的一半。 树形DP的核心:将有向树当作无向树,从根节点结点遍历到叶子节点,从叶子节点回缩到根节点。 思路:一次DFS可求出每个结点的最大子树(不包含其本身)maxn以及子树之和sum(包含其本身),若(n-sum)<=n/2&&maxn<=n/2,则为答案。
#include<cstdio> #include<cstring> #include<algorithm> #include<vector> using namespace std; const int MAXN=10005; vector<int> tree[MAXN]; int n; int mx[MAXN]; int sum[MAXN]; void dfs(int u,int fa) { sum[u]=1; int maxn=0; int s=0; for(int i=0;i<tree[u].size();i++) { int v=tree[u][i]; if(v==fa) continue; dfs(v,u); s+=sum[v]; maxn=max(maxn,sum[v]); } mx[u]=maxn; sum[u]+=s; } int main() { while(scanf("%d",&n)!=EOF) { memset(sum,0,sizeof(sum)); memset(mx,0,sizeof(mx)); for(int i=1;i<=n;i++) tree[i].clear(); for(int i=1;i<=n-1;i++) { int u,v; scanf("%d%d",&u,&v); tree[u].push_back(v); tree[v].push_back(u); } dfs(1,-1); bool flag=false; for(int i=1;i<=n;i++) { if(max(n-sum[i],mx[i])<=n/2) { printf("%d\n",i); flag=true; } } if(!flag) printf("NONE\n"); } return 0; }
相关文章推荐
- 手把手教你怎么免费为微信公众号接入智能客服
- 【CodeForces】A. Twins
- hdu 1565 方格取数(1)(最小割)
- 关于js中的回收机制,通俗版
- (九)利用processing展示简单函数的使用
- 【PAT】1036. 跟奥巴马一起编程(15)
- RxJava过滤操作符 filter
- 计算机网络17--Email应用
- Apache与Tomcat有什么关系和区别
- thinkphp的失败页面(404)
- 【CodeForces】A. Lucky Division
- 学习python之编写简单乘法口诀表实现代码
- js事件
- KO和jQuery (或Prototype等)是竞争关系还是能一起使用?
- python之pymysql模块学习(待完善...)
- thinkphp的调试方案
- [Locked] Largest BST Subtree
- PowerShell-Exchange:Check High Item
- Demo11:content与counter案例
- linux(hadoop)设置网络