P1991 无线通讯网kruskal

Posted 2020-12-12 jason66661010

题目

https://www.luogu.com.cn/problem/P1991

 

 

 分析

本题使用kruskal算法，再能保证所有的节点都能够被连接的情况下，将构成最小生成树的边的长度入队进行排列，除去可以使用卫星电话的节点（就是除去排在前面最长的边）输出最长的边

至于添加边的话，要添加不同节点之间的距离，而且A B 、B A只添加一次就行

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
#define maxm 1000001
#define maxn 1010
struct node
{
    int from;
    int to;
    double dis;
}e[maxm * 2];//自己补的边
struct point
{
    double x;
    double y;

}list[maxn];//点集
bool cmp(struct node &a, struct node &b)
{
    return a.dis < b.dis;
}
bool cmp2 (struct node &a, struct node &b)
{
    return a.dis > b.dis;
}
vector<node> qq;
int father[maxn];
int s,p, cnt = 0;
double allcount = 0;//答案
int find(int x)
{
    if (father[x] == x)
        return x;
    return father[x] = find(father[x]);
}

int kruskal()
{
    sort(e, e + cnt, cmp);
    for (int i = 0; i < cnt; i++)
    {
        int tempx = find(e[i].from);
        int tempy = find(e[i].to);
        if (tempx == tempy)continue;
        qq.push_back(e[i]);
        father[tempx] = tempy;
    }
    return allcount;
}

int main()
{
    scanf("%d%d", &s, &p);
    for (int i = 1; i <= p; i++)
        scanf("%lf%lf", &list[i].x, &list[i].y);
    for (int i = 1; i <= p; i++)father[i] = i;

    for (int i = 1; i <= p; i++)
    {
        for (int j = 1; j <= i; j++)//这里只需要补一半的边 1 2 300 ； 2 1 300 对本题意义一样
        {
            if (i == j)continue;
            e[cnt].from = j; e[cnt].to = i;
            e[cnt].dis = sqrt((list[i].x - list[j].x)*(list[i].x - list[j].x) + (list[i].y - list[j].y)*(list[i].y - list[j].y));

            cnt++;
        }
    }
    kruskal();
    sort(qq.begin(), qq.end(), cmp2);
    double ans = qq[s-1].dis;
    printf("%.2lf",ans );
}