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);

	}

}