Wireless Network POJ

Wireless Network

思路：并查集，如果两个点可以相互传达，就是两个集合的合并，当然，前提是要两个点都被维修过。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>

using namespace std;

#define ll long long
#define pb push_back
#define fi first
#define se second

const int N = 1010;
vector<pair<int ,int > > vp;
int G[N][N];
int fa[N], vis[N];
int n, d;

inline int dis(int x, int y)
{
    return (vp[x].fi - vp[y].fi) * (vp[x].fi - vp[y].fi) + (vp[x].se - vp[y].se) * (vp[x].se - vp[y].se);
} 

void build()
{   
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            G[i][j] = dis(i, j);
        }
    }
}

int Find(int x)
{
    return fa[x] == x ? x : fa[x] = Find(fa[x]);
}

void Union(int x, int y)
{
    int fax = Find(x);
    int fay = Find(y);

    if(fax == fay) return;
    fa[fax] = fay;
}

void solve()
{
    scanf("%d%d", &n, &d);
    d *= d;
    int xx, yy;
    for(int i = 0; i < n; ++i){
        scanf("%d%d", &xx, &yy);
        vp.pb({xx, yy});
    }

    build();

    for(int i = 0; i < n; ++i) fa[i] = i;

    char op[2];
    int x, y;
    while(~scanf("%s", op)){
        if(op[0] == 'O'){
            scanf("%d", &x);
            vis[--x] = 1;//标记该点被修理过
            for(int i = 0; i < n; ++i){
                if(i != x && G[x][i] <= d && vis[i]){
                    Union(x, i);
                }
            }
        }else{
            scanf("%d%d", &x, &y);
            int fax = Find(--x);
            int fay = Find(--y);
            if(fax == fay) printf("SUCCESS
");
            else printf("FAIL
");
        }
    }
}

int main()
{

    solve();

    return 0;
}