poj 3107 树重心

题目大意和思路同上题类似， 本题特殊点为其需要将多组满足要求的结果按递增全部输出。

解题代码：

　　

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
using namespace std;
#define MAX(a,b) (a)>(b)?(a):(b)
#define MIN(a,b) (a)<(b)?(a):(b)
const int N = 50010;

int M[N], n;
int head[N], idx;
struct node{
    int max, sum;
}D[N];
struct Edge{
    int v, next;
}edge[N<<2];

void AddEdge(int a, int b)
{
    edge[idx].v = b; edge[idx].next = head[a]; head[a] = idx++;
    edge[idx].v = a; edge[idx].next = head[b]; head[b] = idx++;
}
void input()
{
    int a, b;    
    //scanf("%d", &n);
    memset( head, 0xff, sizeof(head));
    idx = 0;    
    for(int i = 0; i < n-1; i++)
    {
        scanf("%d%d", &a,&b );
        AddEdge(a, b);
    }
}
int dfs( int u, int pre )
{
    D[u].sum = 1; D[u].max = 0;
    for(int i = head[u]; ~i; i = edge[i].next )
    {
        if( edge[i].v != pre )
        {
            int t = dfs( edge[i].v, u );
            D[u].sum += t;
            D[u].max = MAX( D[u].max, t );
        }
    }
    return D[u].sum;
}
void solve()
{
    dfs( 1, 0 );
    int ans = 0x3fffffff, rt;    
    for(int i = 1; i <= n; i++)
    {
        D[i].max = MAX( D[i].max, n-D[i].sum );    
        if( ans > D[i].max )
        {
            rt = i; ans = D[i].max;    
        }
    }
    bool flag = true;    
    for(int i = 1; i <= n; i++)
    {
        if( D[i].max == ans ){
            if( flag ) flag = false;
            else    printf(" ");
            printf("%d", i);    
        }
    }
}
int main()
{ ;
    while( scanf("%d", &n) != EOF) 
    {
        input();
        solve();
    }
    return 0;
}