HOJ 1811 最小生成樹

最小生成樹入門題，輸入n個座標點，輸出連接這些點的最小距離和。

解法1： Kruskal

//kruskal 1811
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define MAX 105
using namespace std;

struct Point
{
    double x,y;
}point[MAX];

struct Road
{
    int s,d;
    double dis;
}road[MAX*MAX];

double dis(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

bool cmp(Road a,Road b)
{
    return a.dis<b.dis;
}

int father[MAX];

void init()
{
    for(int i=0;i<MAX;i++)
    {
        father[i]=i;
    }
}

int find(int x)
{
    return father[x]==x? x:father[x]=find(father[x]);
}

double kruskal(int index)
{
    double min=0.0;
    int a,b;
    for(int i=0;i<index;i++)
    {
        a=find(road[i].s);
        b=find(road[i].d);
        if(a!=b)
        {
            father[a]=b;
            min+=road[i].dis;
        }
    }
    return min;
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int i,j;
        memset(point,0,sizeof(point));
        for(i=0;i<n;i++)
        {
            scanf("%lf %lf",&point[i].x,&point[i].y);
        }
        int index=0;
        for(i=0;i<n;i++)
        {
            for(j=i+1;j<=n-1;j++)
            {
                road[index].s=i;
                road[index].d=j;
                road[index++].dis=dis(point[i],point[j]);
            }
        }
        sort(road,road+index,cmp);
        init();
        printf("%.2lf\n",kruskal(index));
    }
    return 0;
}

解法2: Prim

//1811 Prim
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define INF  0xffffff
using namespace std;

struct pos
{
    double x,y;
}f[105];

double map[105][105];
bool visited[105];
double dist[105];

double dis(pos a,pos b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

int main()
{
    int n;
    while(scanf(" %d",&n)!=EOF)
    {
        int i,j;
        for(i=0;i<n;i++)
        {
            scanf("%lf %lf",&f[i].x,&f[i].y);
        }
        memset(map,0,sizeof(map));
        memset(visited,0,sizeof(visited));

        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                map[i][j]=dis(f[i],f[j]);
            }
        }
    //memset(dist,100000,sizeof(dist));
    //memset是按字節賦值，不要亂用。。。
       for(int i=0;i<105;i++)
       {
           dist[i]=INF;
       }
        double ans=0;
        dist[0]=0;
        for(int i=0;i<n;i++)
        {
            int k=0;
            double min=INF;
            //找到一個能使和最小的點  將該點加入集合
            for(int j=0;j<n;j++)
            {
                if(!visited[j] && dist[j]<min)
                {
                    min=dist[j];
                    k=j;
                }
            }
            ans+=min;
            visited[k]=1;
            //對錶進行更新
            for(int j=0;j<n;j++)
            {
                if(dist[j]>map[k][j])
                {
                    dist[j]=map[k][j];
                }
            }
        }
        printf("%.2lf\n",ans);
    }
    return 0;
}

解法3: Dijkstra

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define INF  0xffffff
using namespace std;

struct pos
{
    double x,y;
}f[105];

double map[105][105];
bool visited[105];

double dis(pos a,pos b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int i,j;
        memset(f,0,sizeof(f));
        for(i=0;i<n;i++)
        {
            scanf("%lf %lf",&f[i].x,&f[i].y);
        }
        memset(map,INF,sizeof(map));
        memset(visited,0,sizeof(visited));
        for(i=0;i<n-1;i++)
        {
            for(j=i+1;j<n;j++)
            {
                map[i][j]=map[j][i]=dis(f[i],f[j]);
            }
        }
        double ans=0;
        int temp;
        //0爲起點 找點
        visited[0]=1;
        for(int k=0;k<n-1;k++)
        {
            double s=INF;
            //找到一個能使和最小的點  將該點加入集合
            for(int i=1;i<n;i++)
            {
                if(!visited[i] && map[0][i]<s)
                {
                    s=map[0][i];
                    temp=i;
                }
            }
            ans+=s;
            visited[temp]=1;
            //對錶進行更新
            for(int j=1;j<n;j++)
            {
                if(!visited[j] && map[temp][j]<map[0][j])
                {
                    map[0][j]=map[temp][j];
                }
            }
        }
        printf("%.2lf\n",ans);
    }
    return 0;
}