ZOJ 1203 Swordfish MST

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1203

大意：

给出一些点，求MST

把这几天的MST一口气发上来。

kruskal

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN=101;
const int INF=9999999;
int fa[MAXN];

struct point
{
	double x,y;
}ship[MAXN];

struct dot
{
	int x,y;
	double dis;
}dis[MAXN*MAXN];

int find(int cur)
{
	return fa[cur]==cur?cur:fa[cur]=find(fa[cur]);
}

bool operator < (const dot &a,const dot& b)
{
	return a.dis<b.dis;
}

int main()
{
	int n,kase=1;
	while(~scanf("%d",&n),n)
	{
		if(kase!=1)
			printf("\n");
		int i,j;
		for(i=1;i<=n;i++)
			scanf("%lf%lf",&ship[i].x,&ship[i].y);

		
		int len=0;
		for(i=1;i<=n;i++)
		{
			for(j=i+1;j<=n;j++)
			{
				dis[len].x=i;
				dis[len].y=j;
				dis[len].dis=sqrt((ship[j].y -ship[i].y) *(ship[j].y -ship[i].y) + (ship[j].x -ship[i].x)*(ship[j].x -ship[i].x));
				len++;
			}
		}

		for(i=1;i<=n;i++)	
			fa[i]=i;

		sort(dis,dis+len);
		double ans=0;
		for(i=0;i<len;i++)   
		{
			int rootx=find(dis[i].x);
			int rooty=find(dis[i].y);
			if(rootx!=rooty)
			{
				ans+=dis[i].dis;
				fa[rootx]=rooty;
			}
		}
		printf("Case #%d:\n",kase++);
		printf("The minimal distance is: %.2lf\n",ans);
	}
	return 0;

}

prim

#include<cstdio>
#include<cmath>
const int MAXN=101;
const int INF=9999999;
double map[MAXN][MAXN];
double dis[MAXN];
struct point
{
	double x,y;
}ship[MAXN];

void prim(int n)
{
	for(int i=1;i<=n;i++)
		dis[i]=INF;

	bool vis[MAXN]={0};

	int cur=1;            
	vis[cur]=1;
	dis[cur]=0;
	int i,j;


	for(i=1;i<=n;i++)
	{
		double mini=INF;
		for(j=1;j<=n;j++)
			if(!vis[j] && dis[j] > map[cur][j])
				dis[j]=map[cur][j];

		for(j=1;j<=n;j++)
			if(!vis[j] && mini > dis[j])
				mini=dis[cur=j];

		vis[cur]=true;
	}
}
int main()
{
	int n,kase=1;
	while(~scanf("%d",&n),n)
	{
		if(kase!=1)
			printf("\n");
		int i,j;
		for(i=1;i<=n;i++)
			scanf("%lf%lf",&ship[i].x,&ship[i].y);

		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				map[i][j]=	map[j][i] = 
					sqrt((ship[j].y -ship[i].y) *(ship[j].y -ship[i].y) + (ship[j].x -ship[i].x)*(ship[j].x -ship[i].x));
			}
		}

		prim(n);
		double ans=0,ans2=0;//=map[a][b];
		for(i=1;i<=n;i++)
			ans+=dis[i];
		printf("Case #%d:\n",kase++);
		printf("The minimal distance is: %.2lf\n",ans);
	}
}