UVa:10034 Freckles

来源：互联网发布：计算机编程基础编辑：程序博客网时间：2024/06/06 03:18

还是最小生成树的问题。

不过这里的边需要自己求。两个循环就可以求出来。

Kruskal算法：

#include <cstdio>#include <iostream>#include <cstdlib>#include <cmath>#include <cstring>#include <algorithm>using namespace std;int father[105];struct Edge{    int u,v;    double w;};struct Point{    double x,y;};double Distance(Point &a,Point &b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}bool cmp(Edge a,Edge b){    return a.w<b.w;}int find(int x){    return father[x]<0?x:(father[x]=find(father[x]));}int main(){    int T;    scanf("%d",&T);    while(T--)    {        Point p[105];        int n;        scanf("%d",&n);        for(int i=0;i<n;++i)        scanf("%lf%lf",&p[i].x,&p[i].y);        Edge e[10005];        int N=0;        for(int i=0;i<n-1;++i)         for(int j=i+1;j<n;++j)         {             e[N].u=i;             e[N].v=j;             e[N].w=Distance(p[i],p[j]);             N++;         }         sort(e,e+N,cmp);         memset(father,-1,sizeof(father));         double ans=0;         for(int i=0;i<N;++i)         {             Edge t=e[i];             if(find(t.u)!=find(t.v))             {                 father[find(t.u)]=find(t.v);                 ans+=t.w;             }         }         printf("%.2lf\n",ans);         if(T) printf("\n");    }    return 0;}

Prim算法：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#define INF 9999999999using namespace std;int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        double x[105],y[105];        for(int i=0;i<n;++i) scanf("%lf%lf",&x[i],&y[i]);        double e[105][105];        for(int i=0;i<n;++i)         for(int j=0;j<n;++j)         e[i][j]=e[j][i]=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));        double mincost[105];        for(int i=0;i<n;++i) mincost[i]=INF;        bool used[105]={false};        mincost[0]=0;        double res=0;        while(1)        {            int v=-1;            for(int i=0;i<n;++i)        if(!used[i]&&(v==-1||mincost[i]<mincost[v])) v=i;            if(v==-1) break;            used[v]=true;            res+=mincost[v];            for(int i=0;i<n;++i)            mincost[i]=min(mincost[i],e[v][i]);        }        printf("%.2lf\n",res);        if(T) printf("\n");    }    return 0;}