Ultraviolet A 10369-Arctic Network (calculate the K small edge of the Minimum Spanning Tree)

Link:

Http://uva.onlinejudge.org/index.php? Option = com_onlinejudge & Itemid = 8 & category = 24 & page = show_problem & problem = 1310

Question:

Problem C: Arctic Network

The Department of National Defense (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: Every
Outpost will have a radio transceiver and some outposts will in addition have a satellite channel.

Any two outposts with a satellite channel can communicate via the satellite, regardless of their location. Otherwise, two outposts can communicate by radio only if the distance between them does not exceed
D, which depends of the power of the transceivers. higher power yields higher d But costs more. due to purchasing and maintenance considerations, the transceivers at the outposts must be identical; that is, the value of D is the same for every pair of outposts.

Your job is to determine the minimum d Required for the transceivers. There must be at least one communication path (direct or indirect) between every pair of outposts.

The first line of input contains N, the number of test cases. the first line of each test case contains 1 <= S <= 100, the number of satellite channels, and S <p <= 500, the number of outposts. P lines
Follow, giving the (x, y) coordinates of each outpost in km (coordinates are integers between 0 and 10,000 ). for each case, output shocould consist of a single line giving the minimum d required to connect the network. output shoshould be specified to 2 decimal
Points.

Sample Input

12 40 1000 3000 600150 750

Sample output

212.13

Question:

There are n scientific research sites in the south pole. They need to be connected by satellites or radio so that any two sites can be directly or indirectly connected. Any two devices that have satellite devices can communicate with each other directly, regardless of their distance. The distance between the two stations with radio equipment cannot exceed D. D. The longer the cost, the more.

There are now s satellite devices that can be installed, and there are enough radio devices to find a solution, minimize the cost of D (D depends on the path with which all radio communications are consumed ).

Analysis and Summary:

Naturally, we want to find the Minimum Spanning Tree. To minimize the cost of D, we need to make those with the longest paths used to install satellites. S communication satellites can install the longest path of S-1. In the Minimum Spanning Tree, the path length greater than P-S is D.

You only need to find the length of all paths on the minimum generation tree, sort the order, and get the answer by edge.

Code:

1. Kruskal

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define N 505using namespace std;double x[N], y[N], ans[N*N];int n,m,f[N*N],rank[N*N]; struct Edge{    int u,v;    double val;    friend bool operator<(const Edge&a,const Edge&b){        return a.val<b.val;    }}arr[N*N];inline double getDist(double x1,double y1,double x2,double y2){    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}inline void init(){    for(int i=0; i<n*n; ++i)        f[i]=i,rank[i]=0;}int find(int x){    int i,j=x;    while(j!=f[j]) j=f[j];    while(x!=j){        i=f[x]; f[x]=j; x=i;    }    return j;}bool Union(int x, int y){    int a=find(x), b=find(y);    if(a==b)return false;    if(rank[a]>rank[b])        f[b]=a;    else{        if(rank[a]==rank[b])            ++rank[b];        f[a]=b;    }    return true;}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d%d",&m,&n);        init();        for(int i=1; i<=n; ++i)            scanf("%lf%lf",&x[i],&y[i]);        int pos=0;        for(int i=1; i<=n; ++i){            for(int j=i+1; j<=n; ++j)if(i!=j){                arr[pos].u=i, arr[pos].v=j;                arr[pos++].val = getDist(x[i],y[i],x[j],y[j]);            }        }        sort(arr,arr+pos);        int k=0;        for(int i=0; i<pos; ++i){            if(Union(arr[i].u,arr[i].v)){                ans[k++] = arr[i].val;            }        }        printf("%.2f\n", ans[k-m]);    }    return 0;}

2. Prim

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define N 505using namespace std;double x[N], y[N], w[N][N], key[N], ans[N*N];int n,m,pre[N],hash[N]; inline double getDist(double x1,double y1,double x2,double y2){    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}double Prim(){    memset(hash, 0, sizeof(hash));    hash[1] = 1;    for(int i=1; i<=n; ++i){        key[i] = w[1][i]; pre[i] = 1;    }    int k=0;    for(int i=1; i<n; ++i){        int u=-1;        for(int j=1; j<=n; ++j)if(!hash[j]){            if(u==-1 || key[j]<key[u])                 u=j;        }        ans[k++] = key[u];        hash[u] = 1;        for(int j=1; j<=n; ++j)if(!hash[j]&&key[j]>w[u][j]){            key[j]=w[u][j]; pre[j]=u;        }    }    sort(ans, ans+k);    return ans[k-m];}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d%d",&m,&n);        for(int i=1; i<=n; ++i)            scanf("%lf%lf",&x[i],&y[i]);        int pos=0;        memset(w, 0, sizeof(w));        for(int i=1; i<=n; ++i){            for(int j=i+1; j<=n; ++j){                w[i][j]=w[j][i]=getDist(x[i],y[i],x[j],y[j]);            }        }         printf("%.2f\n", Prim());    }    return 0;}

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  Original
  Http://blog.csdn.net/shuangde800
  , By d_double (reprinted, please mark)