Skip to main content

Jogging Trails

                                      Jogging Trails

                                                       bitwise DP 


#include <iostream>
#include <cstring>
#include <climits>
#include <cstdio>

#define INF INT_MAX
#define sf(a) scanf("%d", &a)

#define set(b, pos) ( b | (1 << pos) )
#define off(b, pos) ( b & ~( 1 << pos) )
#define test(b, pos) (b & (1 << pos) )

using namespace std;

int v, e, st, en, wt, weight, bits;

int graph[15][15];
int deg[15]; //stores degree of a node
int memo[1 << 15];

int solve(int bits)
{
    if(memo[bits] >= 0)
        return memo[bits];

    int ans = INF;
    int b;

    for(int i = 0; i < v; i++)
    {
        for(int j = i + 1; j < v; j++)
        {
            if(test(bits, i) && test(bits, j))
            {
                b = bits;
                b = off(b, i);
                b = off(b, j);

                ans = min(ans, solve(b) + graph[i][j]);
            }
        }
    }

    return memo[bits] = ans;
}

//finds the shortest distance between every node
void floyd()
{
    for(int k = 0; k < v; k++ )
    {
        for(int i = 0; i < v; i++ )
        {
            for(int j = 0; j < v; j++ )
            {
                if(graph[i][k] != INF && graph[k][j] != INF)
                    graph[i][j] = min(graph[i][k]+graph[k][j], graph[i][j]);
            }
        }
    }
}


int main()
{
    while(scanf("%d %d,", &v, &e) == 2)
    {
        for(int i = 0; i < v; i++)
        {
            for(int j = 0; j < v; j++)
            {
                graph[i][j] = INF;
            }
        }

        memset(deg, 0, sizeof(deg));
        weight = 0;
        bits = 0;

        for(int i = 0; i < e; i++)
        {
            sf(st); sf(en); sf(wt);

            st--, en--;

            graph[st][en] = graph[en][st] = min(graph[st][en], wt);

            weight += wt;

            deg[st]++, deg[en]++;
        }

        floyd();

        //sets 1 for each node that has odd degree
        for(int i = 0; i < v; i++)
        {
            if(deg[i] & 1)
            {
                bits = set(bits, i);
            }
        }

        memset(memo, -1, sizeof(memo));
        memo[0] = 0;

        weight += solve(bits);

        cout << weight << endl;
    }

    return 0;
}

Comments

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++)  

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; } }

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