【POJ 1655】Balancing Act

Balancing Act

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18593		Accepted: 7849

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

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题解：本应该快乐的树形DP，和标程没啥区别，甚至连oo都改了差不多了

我就是超时，标程对的，无语，不知道哪错啦。

正解：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAX=20005;

int visit[MAX],num[MAX],head[MAX],ans;
int n,mark,k,sum,sizemin;

struct{int v,next;}e[MAX*2];

void  addedge(int u,int v)
{
    k++;
    e[k].v=v;
    e[k].next=head[u];
    head[u]=k;
    return;
}

void dfs(int u)
{// cout<<"L"<<endl;
    int i,t,min;
    visit[u]=1;
    num[u]=1;min=-1;
    for(i=head[u];i;i=e[i].next)
    {
        t=e[i].v;
        if(!visit[t])
        {
            dfs(t);
            num[u]+=num[t];
            min=max(min,num[t]);
        }
    }
    min=max(min,n-num[u]);
    if(min<sizemin)
    {
        ans=min;
        mark=u;
        sizemin=min;
    }

   return ;
}

int main()
{
    int CASE,u,v,i;
    scanf("%d",&CASE);
    while(CASE--)
    {
        scanf("%d",&n);
        k=0;
        memset(head,0,sizeof(head));
        for(i=1;i<n;i++)
        {
           scanf("%d%d",&u,&v);
           addedge(u,v);
           addedge(v,u);
        }
 // cout<<"no"<<endl;
        memset(num,0,sizeof(num));
        memset(visit,0,sizeof(visit));
        k=0;
        sizemin=MAX+10;
        dfs(1);
        printf("%d %d
",mark,ans);


    }
    return 0;
}

 我的代码，找了一个小时还是错，。。。。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
typedef long long ll;
using namespace std;
const int N=40005;
const int oo=0x3f3f3f3f;
int ycll,n,cnt,head[N],vis[N],num[N],mn,ans,mark; 
struct node{
    int v;
    int next;
}e[N]; 
void add(int u,int v){
    cnt++; e[cnt].v=v;
    e[cnt].next=head[u];
    head[u]=cnt;
}
void dfs(int u){
    vis[u]=1;
    num[u]=1;
    int jjj=-1;
    for(int i=head[u];i;i=e[i].next){
        int dv=e[i].v;
        if(!vis[dv]){
            dfs(dv); num[u]+=num[dv];
            jjj=max(jjj,num[dv]);
        }
    }
    jjj=max(jjj,n-num[u]);
    if(jjj<mn) 
       { ans=jjj; mark=u; mn=oo; }
}
int main(){
    freopen("1655.in","r",stdin);
    freopen("1655.out","w",stdout);
    scanf("%d",&ycll);
    while(ycll--){
        scanf("%d",&n);
        cnt=0; 
        memset(num,0,sizeof(num));
        memset(vis,0,sizeof(vis));
        memset(head,0,sizeof(head));
        int x,y;
        for(int i=1;i<n;i++){
            scanf("%d %d",&x,&y);
            add(x,y); add(y,x);
        }
        cnt=0; mn=N+10; dfs(1);
        printf("%d %d
",mark,ans);
    }
    return 0;
}