POJ 3099 Go Go Gorelians

http://poj.org/problem?id=3099

树的重心：找到一个点,其所有的子树中最大的子树节点数最少,那么这个点就是这棵树的重心

求树的重心

如何在点中构造符合条件的树

得到树后 从任意一个点出发 dfs一次找到离这个点最远的点作为root1

在以root1出发 同样的方式求得root2

root1-root2就是这个树的直径 那么中心必然在这个直径上， 如果直径上点的个数是奇数的话那么 这个点就是重心 否则有两个重心 分别输出

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <vector>
#include <math.h>
#define MAXN 1007

using namespace std;

int dp[MAXN];
int id[MAXN];
int pre[MAXN];
double dist[MAXN][MAXN];
double x[MAXN], y[MAXN], z[MAXN];
int path[MAXN];
int n;
vector<int> G[MAXN];

double getdis(int i, int j)
{
    double dx = x[i] - x[j];
    double dy = y[i] - y[j];
    double dz = z[i] - z[j];
    return sqrt(dx*dx+dy*dy+dz*dz);
}

void build()
{
    for(int i = 0; i < n; i++)
    {
        double dis = 0x3fffffff;
        int obj = -1;
        for (int j = 0; j < i; j++)
        {
            if (dist[id[i]][id[j]] < dis)
            {
                dis = dist[id[i]][id[j]];
                obj = j;
            }
        }
        if (obj == -1) continue;
        int u = id[i], v = id[obj];
        G[u].push_back(v);
        G[v].push_back(u);
    }
}
void dfs(int x, int par, int len)
{
    dp[x] = len;
    for (int i = 0; i < G[x].size(); i++)
    {
        if (G[x][i] != par)
            dfs(G[x][i], x, len+1);
    }

}


void dfs1(int x, int par, int len)
{
    dp[x] = len;
    pre[x] = par;
    for (int i = 0; i < G[x].size(); i++)
    {
        if(G[x][i] != par)
            dfs1(G[x][i], x, len+1);
    }
}
int main()
{
    while (~scanf("%d", &n))
    {
        if (n == 0) break;
        memset(dp, 0, sizeof(dp));
        memset(path, 0, sizeof(path));
        memset(dist, 0, sizeof(dist));
        memset(pre, 0, sizeof(pre));
        for (int i = 0; i < MAXN; i++)
            G[i].clear();
        for (int i = 0; i < n; i++)
        {
            scanf("%d%lf%lf%lf", &id[i], &x[i], &y[i], &z[i]);
        }
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                dist[id[i]][id[j]] = getdis(i, j);
        build();
        dfs(id[0], -1, 0);//权值都为1
        int tmp = 0;
        //找到离节点最远的节点 得到root1
        int root1, root2;
        for (int i = 0; i < n; i++)
        {
            if ( dp[id[i]] > tmp)
            {
                tmp = dp[id[i]];
                root1 = id[i];
            }
        }
        memset(dp, 0, sizeof(dp));
        dfs1(root1, -1, 0);
        tmp = 0;
        for (int i = 0; i < n; i++)
        {
            if (dp[id[i]] > tmp)
            {
                tmp = dp[id[i]];
                root2 = id[i];
            }
        }//找到root2;
        int s = 0;
        for(int i = root2; i != -1; i = pre[i])
        {
            path[++s] = i;
        }
        int mid = (s+1)/2;
        if (s % 2 == 1)
        {
            printf("%d
", path[mid]);
        }
        else
        {
            int a = path[mid], b = path[mid+1];
            printf("%d %d
", min(a, b), max(a, b));
        }

    }

    return 0;
}

关于树的重心更多的内容

转载 ：http://www.cnblogs.com/patrickzhou/p/5867208.html