蓝桥杯省赛模拟：村庄建设、郊外植树。

村庄建设

题目概述

补题链接

村庄建设

方法和代码

最小生成树，但是要自己构边，可以用prime，也可以用kruskra，这里用kruskra来写。

/*
    Code by lifehappy 2020:04:19
    方法：Kruskra最小生成树
*/
#include<bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const int N = 1e6 + 10;
int sgn(double x) {
    if(fabs(x) < eps)   return 0;
    if(x > 0)   return 1;
    return -1;
}
struct point {
    double x, y, z;
    point(double a = 0.0, double b = 0.0, double c = 0.0) : x(a), y(b), z(c) {} 
    point operator - (point t) {
        return point(x - t.x, y - t.y, z - t.z);
    }
}p[N];
double dis(point a, point b) {
    a = a - b;
    return sqrt(a.x * a.x + a.y * a.y) + a.z * a.z;
}

struct Edge {
    int x, y;
    double dis;
    bool operator < (Edge t) {
        return (sgn(dis - t.dis) <= 0);
    }
}a[N];
int n, cnt, fa[N], sum;
void init() {
    for(int i = 0; i < n; i++)
        fa[i] = i;
}
int find(int rt) {
    return rt == fa[rt] ? rt : fa[rt] = find(fa[rt]);
}
int main() {
    // freopen("in.txt", "r", stdin);
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
        scanf("%lf %lf %lf", &p[i].x, &p[i].y, &p[i].z);
    for(int i = 0; i < n; i++) {
        for(int j = i + 1; j < n; j++) {
            a[cnt].x = i, a[cnt].y = j;
            a[cnt].dis = dis(p[i], p[j]);
            cnt++;
        }
    }
    init();
    double ans = 0;
    sort(a, a + cnt);
    for(int i = 0; i < cnt; i++) {
        int fx = find(a[i].x);
        int fy = find(a[i].y);
        if(fx != fy) {
            fa[fx] = fy;
            ans += a[i].dis;
            sum++;
        }
        if(sum + 1 == n)    break;
    }
    printf("%.2f
", ans);
    return 0;
}

郊外植树

题目概述

补题链接

郊外植树

方法和代码

贪心的思想，优先栽植半径大的数。

/*
    Code by lifehappy 2020:04:19
    方法：贪心
*/
#include<bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const int N = 50;
int sgn(double x) {
    if(fabs(x) < eps)   return 0;
    if(x > 0)   return 1;
    return -1;
}
struct point {
    double x, y, r;
    point(double a = 0.0, double b = 0.0, double c = 0.0) : x(a), y(b), r(c) {}
    point operator - (point t) {
        return point(x - t.x, y - t.y, r - t.r);
    }
    bool operator < (point t) {
        return (sgn(r - t.r)  >= 0);
    }
}a[N], b[N];
double dis(point a, point b) {
    a = a - b;
    return sqrt(a.x * a.x + a.y * a.y);
}

int n;
int main() {
    // freopen("in.txt", "r", stdin);
    cin >> n;
    for(int i = 0; i < n; i++)
        cin >> a[i].x >> a[i].y >> a[i].r;
    sort(a, a + n);
    int ans = 0, pos = 0;
    for(int i = 0; i < n; i++) {
        int flag = 1;
        for(int j = 0; j < pos; j++) {
            double d = dis(a[i], b[j]);
            if(sgn(d - a[i].r - b[j].r) == -1) {
                flag = 0;
                break;
            }
        }
        if(flag) {
            ans += a[i].r * a[i].r;
            b[pos++] = a[i];
        }
    }
    cout << ans << endl;
    return 0;
}