[蓝桥杯练习]通电-编程知识

[蓝桥杯练习]通电

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kruskal做法(加边)

#include <bits/stdc++.h>
using namespace std;
int x[10005],y[10005],z[10005];//存储i点的x与y坐标 
int bcj[10005];//并查集 
struct Edge{//边 int v1,v2; double w;
}edge[2000005];
int cmp(Edge a, Edge b){return a.w < b.w;}
int find(int x){//并查集查找 if(bcj[x]!=x)bcj[x]=find(bcj[x]);//带路径压缩return bcj[x];
}
void merge(int v1,int v2){//并查集合并 v1=find(v1);v2=find(v2);bcj[v2]=v1;
}
int main(){//int n;cin>>n;for(int i=1;i<=n;++i)bcj[i]=i;//并查集初始化//for(int i=1;i<=n;i++)cin>>x[i]>>y[i]>>z[i];//读取结点x坐标y坐标 int cnt=0;for(int v1=1;v1<=n;++v1)for(int v2=v1+1;v2<=n;++v2){//求出任意两点的权重 double w=sqrt(pow((x[v1]-x[v2]),2)+pow((y[v1]-y[v2]),2))+pow((z[v1]-z[v2]),2);//v1的x坐标减去v2的x坐标... edge[++cnt]={v1,v2,w};}sort(edge+1,edge+cnt+1,cmp);int MSTm = 0;double sumw = 0.0;//解决了任意孤立点(一个点就是一个集合),然后对每个点作n-1个点的相连边并排序,保证每次取得是最短的,并且使用并查集避免出现环路. //取得边是离散的,总可以取完 for(int i=1;i<=cnt;++i){//依次取得边,其权重递增 if(find(edge[i].v1)!=find(edge[i].v2)){//若两边端点不属于同一个集合,则合并 merge(edge[i].v1,edge[i].v2);//每次取一条边并将其标记为一个集合,使其不出现环路 ++MSTm;sumprimzuofaw+=edge[i].w;//获取MST中最大的边 }if(MSTm==n-1)break;//n-1条边就可以构造MST }cout<<fixed<<setprecision(2)<<sumw<<endl;return 0;
}

prim做法(加点)

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 1005;
vector<int> demo;
double closest[MAXN], lowcost[MAXN];
int m, n;             // m为节点的个数，n为边的数量
double G[MAXN][MAXN]; // 邻接矩阵
double prim()
{for (int i = 0; i < m; i++){lowcost[i] = INF;}for (int i = 0; i < m; i++){closest[i] = 0;}closest[0] = -1;             // 加入第一个点，-1表示该点在集合U中，否则在集合V中int num = 0,  e = 0; // e为最新加入集合的点double ans=0;while (num < m - 1)          // 加入m-1条边{int miedge = -1;double micost = INF;for (int i = 0; i < m; i++)if (closest[i] != -1){double temp = G[e][i];if (temp < lowcost[i]){lowcost[i] = temp;closest[i] = e;}if (lowcost[i] < micost)micost = lowcost[miedge = i];}ans += micost;demo.push_back(micost);closest[e = miedge] = -1;num++;}return ans;
}struct node
{double x, y, h;
} dis[MAXN];double getDistance(node a, node b)
{return sqrt(pow(a.x - b.x, 2) + pow(a.y - b.y, 2)) + pow(a.h - b.h, 2);
}int main()
{scanf("%d", &m);for (int i = 0; i < m; i++)scanf("%lf%lf%lf", &dis[i].x, &dis[i].y, &dis[i].h);for (int i = 0; i < m - 1; i++)for (int j = i + 1; j < m; j++){G[i][j] = getDistance(dis[i], dis[j]);G[j][i] = G[i][j];}printf("%.2lf", prim());// for (int i = 0; i < m - 1; i++)//     cout << demo[i] << " ";return 0;
}