【POJ 1655】 Balancing Act

【题目链接】

点击打开链接

【算法】

树形DP求树的重心

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 20010

int i,T,n,u,v,ans,pos;
vector<int> E[MAXN];
int size[MAXN];

template <typename T> inline void read(T &x) {
        int f = ; x = ;
        char c = getchar();
        for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
        for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
        x*= f;
}

template <typename T> inline void write(T x) {
    if (x < ) { putchar('-'); x = -x; }
    if (x > ) write(x/);
    putchar(x%+'');
}

template <typename T> inline void writeln(T x) {
    write(x);
    puts("");
}

inline void dfs(int x,int fa) {
        int i,son,m=;
        size[x] = ;
        for (i = ; i < E[x].size(); i++) {
                son = E[x][i];
                if (son != fa) {
                        dfs(E[x][i],x);
                        size[x] += size[son];
                        if (size[son] > m) m = size[son];
                }
        }
        if (n - size[x] > m) m = n - size[x];
        if (m < ans) {
                ans = m;
                pos = x;
        }
}

int main() {

        read(T);
        while (T--) {
                read(n);
                for (i = ; i <= n; i++) E[i].clear();
                for (i = ; i < n; i++) {
                        read(u); read(v);
                        E[u].push_back(v);
                        E[v].push_back(u);
                }
                ans = 2e9;
                dfs(,);
                printf("%d %d\n",pos,ans);
        }

        return ;

}