洛谷P1991题解

要求最大值最小，不难想到使用二分法。我们每次二分一个“最大的D”，然后将大于D的边舍去，然后看一下去边后的图有多少个连通分量。连通分量的数目就是增配卫星电话的数目。

至于求连通分量数目我们可以使用并查集。

#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;

#define MaxP 505

int S, P;
struct NODE{
    double x, y;
}node[MaxP];

struct EDGE{
    int x, y, f;
    double w;
}edge[MaxP*MaxP];
int m;

bool cmp(EDGE a, EDGE b){
    return a.w < b.w;
}

double distance(int x, int y){
    double tmp = (double)(node[x].x-node[y].x)*(node[x].x-node[y].x);
    tmp += (double)(node[x].y-node[y].y)*(node[x].y-node[y].y);
    tmp = sqrt(tmp);
    return tmp;
}

int fa[MaxP];
int find(int x){
    return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);
}
bool flag[MaxP];

bool check(int d){
    for(int i = 1; i<=P; ++i) fa[i] = i;
    for(int i = 1; i<=d; ++i){
        int fx = find(edge[i].x);
        int fy = find(edge[i].y);
        if(fx == fy) continue;
        fa[fx] = fy;
    }
    for(int i = 1; i<=P; ++i) flag[i] = false;
    for(int i = 1; i<=P; ++i) flag[find(i)] = true;
    int cnt = 0;
    for(int i = 1; i<=P; ++i) if(flag[i]) ++cnt;
    if(cnt<=S) return true;
    return false;
}

int main(){
    scanf("%d %d", &S, &P);
    for(int i = 1; i<=P; ++i) scanf("%lf %lf", &node[i].x, &node[i].y);
    for(int i = 1; i<=P; ++i)
        for(int j = i+1; j<=P; ++j){
            edge[++m].x = i;
            edge[m].y = j;
            edge[m].w = distance(i, j);
        }
    sort(edge+1, edge+1+m, cmp);
    int l = 1, r = m, mid, ans;
    while(l<=r){
        mid = (l+r)>>1;
        if(check(mid)) ans = mid, r = mid-1;
        else l = mid+1;
    }
    printf("%.2lf\n", edge[ans].w);
    return 0;
}