• SPOJ PT07Z, Longest path in a tree 树的直径


     SPOJ PT07Z, Longest path in a tree 

    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <string>
    #include <set>
    #include <queue>
    #include <map>
    #include <sstream>
    #include <cstdio>
    #include <cstring>
    #include <numeric>
    #include <cmath>
    #include <iomanip>
    #include <deque>
    #include <bitset>
    #include <cassert>
    //#include <unordered_set>
    //#include <unordered_map>
    #define ll              long long
    #define pii             pair<int, int>
    #define rep(i,a,b)      for(int  i=a;i<=b;i++)
    #define dec(i,a,b)      for(int  i=a;i>=b;i--)
    #define forn(i, n)      for(int i = 0; i < int(n); i++)
    using namespace std;
    int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
    const long long INF = 0x7f7f7f7f7f7f7f7f;
    const int inf = 0x3f3f3f3f;
    const double pi = acos(-1.0);
    const double eps = 1e-6;
    const int mod = 1e9 + 7;
    
    inline ll read()
    {
        ll x = 0; bool f = true; char c = getchar();
        while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
        while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
        return f ? x : -x;
    }
    inline ll gcd(ll m, ll n)
    {
        return n == 0 ? m : gcd(n, m % n);
    }
    void exgcd(ll A, ll B, ll& x, ll& y)
    {
        if (B) exgcd(B, A % B, y, x), y -= A / B * x; else x = 1, y = 0;
    }
    inline int qpow(int x, ll n) {
        int r = 1;
        while (n > 0) {
            if (n & 1) r = 1ll * r * x % mod;
            n >>= 1; x = 1ll * x * x % mod;
        }
        return r;
    }
    inline int inv(int x) {
        return qpow(x, mod - 2);
    }
    ll lcm(ll a, ll b)
    {
        return a * b / gcd(a, b);
    }
    /**********************************************************/
    const int N = 1e4 + 5;
    int d[N], c;
    void dfs1(int x, int fa, vector<vector<int>>& a)
    {
        for (int i = 0; i < a[x].size(); i++)
        {
            int to = a[x][i];
            if (to != fa)
            {
                d[to] = d[x] + 1;
                if (d[to] > d[c])
                    c = to;
                dfs1(to, x, a);
            }
        }
    }
    
    int main() {
        ios::sync_with_stdio(false);cin.tie(nullptr);
        int n;
        cin >> n;
        vector<vector<int>> a(n + 1);
        rep(i, 1, n-1)
        {
            int u, v;
            cin >> u >> v;
            a[u].push_back(v);
            a[v].push_back(u);
        }
        d[1] = 1;
        dfs1(1, -1, a);
        memset(d, 0, sizeof(d));
        d[c] = 1;
        dfs1(c, -1, a);
        cout << d[c] - 1 << endl;
    }
  • 相关阅读:
    Python for Data Science
    Python for Data Science
    Python for Data Science
    团队项目选题报告(I know)
    结对第一次作业——原型设计
    团队展示(I know)
    软件工程实践第二次作业——个人项目实战(数独)
    软件工程实践第一次作业--准备篇
    Java微笔记(9)
    Java微笔记(8)
  • 原文地址:https://www.cnblogs.com/dealer/p/13417678.html
Copyright © 2020-2023  润新知