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：

　　

View Code

#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;

}