POJ-1655 Balancing Act 树的重心

题意:完全符合树的重心:即找到一个点,其所有的子树中最大的子树节点最少.

代码如下:

#include <cstdlib>
#include <cstring>
#include <cstdio> 
#include <iostream>
#include <algorithm>
#define MAXN 20000
using namespace std;

//说白了这一题就是求一棵树的重心 

int N, idx; // 说明有N个节点
//由于节点较多,且数为稀疏图,因此采用邻接表的形式来存储

struct Node {
    int x, next, cnt; // cnt用来说明该边所连接节点为子树的总节点数 
    bool vis;
}e[(MAXN<<1)+5]; // 双向边所致 

int head[MAXN+5];

void addedge(int a, int b) {
    ++idx;
    e[idx].vis = false, e[idx].cnt = 0;
    e[idx].x = b, e[idx].next = head[a];
    head[a] = idx;
}

int dfs(int x) { // 得到某条边连接的子树的节点数(入口边不进行计算,否则全为N了)
    int sum = 0;
    for (int i = head[x]; i != -1; i = e[i].next) {
        if (!e[i].vis) { // 如果这条边没有被访问过
            e[i].vis = e[i^1].vis = true;
            e[i].cnt += dfs(e[i].x);
            e[i^1].cnt = N-e[i].cnt;
        }
        sum += e[i].cnt;
    }
    return sum + 1;
}

void solve(int &x, int &y) {
    y = 0x7fffffff;
    for (int i = 1; i <= N; ++i) {
        int Max = 0x7fffffff+1;
        for (int k = head[i]; k != -1; k = e[k].next) {
            Max = max(Max, e[k].cnt);
        }
        if (Max < y) {
            y = Max;
            x = i;
        }
    }
}

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        int a, b, x, y;
        memset(head, 0xff, sizeof (head));
        idx = -1;
        scanf("%d", &N);
        for (int i = 1; i < N; ++i) {
            scanf("%d %d", &a, &b);
            addedge(a, b);
            addedge(b, a);
        }
        dfs(1);
        solve(x, y);
        printf("%d %d\n", x, y);
    }
    return 0;
}