Interligando a Sildavia

Hi, it's been a time since I post something useful in the blog, today I'll talk about a problem from the 'Maratona Mineira de Programação' as the statement is in Portuguese, and probably anyone who's interested to solve it is a native Portuguese speaker, it make more sense to write the post in Portuguese, but to keep my English in practice, I'll write in English:

The statement and Example test cases can be found here: http://br.spoj.pl/problems/INTERLMG/

The problem asks to return the minimum length of a string to connect all the pair of vertexes:

I used the follow strategy, which is a little bit slow, but it's enough to pass in the system test:

Think the each point as a vertex in a graph, and the string pieces, the weight of the edges between them, we want to find the minimum possible value to the weight of all edges(which have to be equal) to connect all the points;

I've ran a binary search, looking for the minimum weight for this edges, as the number can be real, I used a bound(in this case, 50) to doesn't exceed the time limit, and used each point as a vertex in a graph, checking as a edge if the length of the strihere comes the code:

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

using namespace std;

int n, vis[1010], path[1010][1010];
double x[1010], y[1010];

void dfs(int a) {
    vis[a] = true;
    for(int i = 0; i < n; i++) {
        if(path[a][i] == 1 && !vis[i]) {
            dfs(i);
        }
    }
}

int main(void) {
    ios::sync_with_stdio(false);
    while(cin >> n && n) {
        for(int i = 0; i < n; i++) {
            cin >> x[i] >> y[i];
        }
        double low = 0.0, high = 10000000.0, mid;
        for(int z = 0; z <= 50; z++) {
            memset(vis, 0, sizeof(vis));
            bool has = false;
            mid = low + (high - low) / 2;
            for(int i = 0; i < n; i++) {
                for(int j = i + 1; j < n; j++) {
                    if(hypot(x[i] - x[j], y[i] - y[j]) < mid) {
                        path[i][j] = path[j][i] = 1;
                    } else {
                        path[i][j] = path[j][i] = 0;
                    }
                }
            }
            dfs(0);
            for(int i = 0; i < n; i++) if(vis[i] == 0) has = true;
            if(has) {
                low = mid;
            } else {
                high = mid;
            }
        }
        printf("%.4f\n", low);
    }
    return 0;
}

Then, I realized that it's also possible to solve the problem in a better way O(N^2 + (N log(N)) + M) with N as the number of points, and M as the edges between the points, here comes the code again:

#include <iostream>
#include <vector>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

using namespace std;

int n, vis[1010], path[1010][1010];
double x[1010], y[1010];

struct UnionFind {
    int N, *id, *sz;

    UnionFind(int _N) {
        id = new int[_N];
        sz = new int[_N];
        for(int i = 0; i < _N; i++) {
            id[i] = i;
            sz[i] = 1;
        }
        N = _N;
    }
    int root(int i) {
        while(i != id[i]) {
            id[i] = id[id[i]];
            i = id[i];
        }
        return i;
    }
    bool find(int p, int q) {
        return root(p) == root(q);
    }
    void unite(int p, int q) {
        int i = root(p);
        int j = root(q);
        if(i == j) return;
        if(sz[i] < sz[j]) {
            id[i] = j; sz[j] += sz[i];
        } else {
            id[j] = i; sz[i] += sz[j];
        }
    }
};

struct edge {
    int a, b;
    double len;

    edge(){}
    edge(int _a, int _b, double _len) {
        a = _a;
        b = _b;
        len = _len;
    }
    bool operator<(const edge &e) const {
        return len < e.len;
    }
};


int main(void) {
    ios::sync_with_stdio(false);
    while(cin >> n && n) {
        vector<edge> vs;
        for(int i = 0; i < n; i++) {
            cin >> x[i] >> y[i];
        }
        for(int i = 0; i < n; i++) {
            for(int j = i + 1; j < n; j++) {
                vs.push_back(edge(i, j, (double)hypot(x[i] - x[j], y[i] - y[j])));
            }
        }
        UnionFind uf(n+1);
        sort(vs.begin(), vs.end());
        double ans = 0.0;
        for(int i = 0; i < vs.size(); i++) {
            if(!uf.find(vs[i].a, vs[i].b)) {
                uf.unite(vs[i].a, vs[i].b);
                if(vs[i].len > ans) {
                    ans = vs[i].len;
                }
            }
        }
        printf("%.4f\n", ans);
    }
    return 0;
}