Skip to main content

Civilization

                                           Civilization


#include <iostream>
#include <vector>
#include <stdio.h>
#define sf(i) scanf("%d", &i)
#define pf(i) printf("%d\n", i)
#define pb push_back
using namespace std;

const int N = 3e5 + 1;
vector<int> g[N];
int parent[N];
int diameter[N];
bool isUsed[N];
int n, m, q, maxi, maxiv;

int rad(int a)
{
    return (diameter[a] + 1) / 2;
}
int find(int a)
{
    if(parent[a] == a)
    {
        return a;
    }
    return parent[a] = find(parent[a]);
}

void unite1(int a, int b)
{
    a = find(a);
    b = find(b);
    if(a > b)
        swap(a, b);
    parent[b] = a;
}

void unite(int a, int b)
{
    a = find(a);
    b = find(b);
    if(a > b)
    {
        swap(a, b);
    }

    parent[b] = a;

    diameter[a] = max(rad(a) + rad(b) + 1, max(diameter[a], diameter[b]));

}

void dfs(int v, int depth, int p)
{
    if(depth > maxi)
    {
           maxi = depth;
           maxiv = v;
    }
    for(int i = 0; i < g[v].size(); i++)
    {
        if(g[v][i] == p)
            continue;
        dfs(g[v][i], depth + 1, v);
    }
}
int findLongestWay(int i)
{
    maxi = -1;
    dfs(i, 0, -1);
    maxi = -1;
    dfs(maxiv, 0, -1);
    return maxi;
}
int main()
{
    int a, b, x, y, num;
    sf(n), sf(m), sf(q);

    for(int i = 0; i <= n; i++)
    {
        parent[i] = i;
        diameter[i] = 0;
        isUsed[i] = false;
    }

    for(int i = 0; i < m; i++)
    {
        sf(a), sf(b);
        g[a].pb(b);
        g[b].pb(a);
        if(find(a) != find(b))
            unite1(a, b);
    }

    for(int i = 1; i <= n; i++)
    {
        int p = find(i);
        if(!isUsed[p])
        {
            isUsed[p] = true;
            diameter[p] = findLongestWay(i);
        }
    }

    for(int i = 0; i < q; i++)
    {
        sf(num);
        if(num == 1)
        {
            sf(x);
            x = find(x);
            pf(diameter[x]);
        }
        else
        {
            sf(x), sf(y);
            if(find(x) != find(y))
                unite(x, y);
        }

    }
    return 0;
}

Comments

Post a Comment

Popular posts from this blog

MAIN72

MAIN72 - Subset sum #include <iostream> #include <cstring> using namespace std; bool dp[100001][1001]; int arr[1001]; int main() {     int t, n;     long long int sum;     cin >> t;     while(t--)     {         cin >> n;         memset(dp, false, sizeof(dp));         sum = 0;         for(int i = 0; i < n; i++)         {             cin >> arr[i];             sum += arr[i];         }         for(int i = 0; i < n; i++)             dp[0][i] = true; // 0 sum         for(int i = 1; i < n; i++)             dp[i][0] = false; // sum is i but 0 element         for(long int i = 1; i <= sum; i++)         {             for(int j = 1; j <= n; j ++)             {                 dp[i][j] = dp[i][j - 1];                 if(i >= arr[j - 1])                     dp[i][j] = dp[i][j] || dp[i - arr[j - 1]][j - 1];             }         }         long int result = 0;         for(int i = 1; i <= sum; i++)  
Post a Comment
Read more

War of XORs- XORIER

Problem Link #include <iostream> using namespace std; int main() { int t, n, odd, even; cin >> t; while(t--) { cin >> n; int i,arr[n],freq[1100001]={0}; long res = 0; odd = even = 0; for(int i = 0; i < n; i++) { cin >> arr[i]; freq[arr[i]]++; } for(int i = 0; i < n; i++) { if(arr[i] & 1) { odd++; } else { even++; } } for(int i = 0; i < n; i++) { if(arr[i] % 2) { res += odd; } else { res += even; } res -= freq[arr[i] ^ 2]; res -= freq[arr[i]]; } cout << res / 2 << endl; } }
Post a Comment
Read more

GCJ101BB - Picking Up Chicks

Problem Link /* explanation     lets solve the problem only for 2 chicken.     s[i] = speed of chicken i     pos[i] = position of chicken i     if s[i] > s[i - 1] then no problem, just check whether both can reach b within time or not.     if s[i] < s[i - 1] then there is a chance that i can slow down i - 1.     lets say s[i] = 1 m/sec and s[i - 1] = 2 m/sec and time limit is T and point to reach is B.     for s[i] pos[i] can be at max B - T. if pos is greater than B-T it can not reach within Tsec.     and for s[i - 1] pos[i - 1] can be at max (B-T)*2. if pos[i - 1] > (B-T)*2 it can not reach within Tsec.     at T sec i will be at B and i - 1 will also be at B. at T - 1 i will be at B-T-1 and i-1 will be at B-T-2 and so on. as we can see i -1 will always be behind i. so there will not be any collision.     if i is pos[i] < B-T then i can reach B before T sec and it will not cause any problem .     problem may occur if pos[i - 1]< (B -T)*2 as it can me
Post a Comment
Read more