HDOJ1162（Prim 或 Kruscal）

来源：互联网发布：rsa密码算法编辑：程序博客网时间：2024/06/06 04:56

标准最小生成树

本文提供两种做法：

Prim
Kruscal

Prim解法：

#include<iostream>#include<math.h>using namespace std;struct tag_point{double x,y;}point[105];double map[105][105];int N;bool judge[105];double dis(int i,int j){return sqrt((point[i].x-point[j].x)*(point[i].x-point[j].x)+(point[i].y-point[j].y)*(point[i].y-point[j].y));}double Prim(){int i,j,tnum,k;double sum=0,tmin;//初始化判断数组for(i=1;i<=N;i++) judge[i]=false;judge[1]=true;//Primfor(k=1;k<N;k++){tmin=0x3f3f3f3f;tnum=0;for(i=1;i<=N;i++){for(j=2;j<=N;j++){if(!judge[j]&&judge[i])if(map[i][j]<tmin){tmin=map[i][j];tnum=j;}//end if }//end j}// end ijudge[tnum]=true;sum+=tmin;}//end kreturn sum;}int main(){int i,j;//freopen("1.txt","r",stdin);while(cin>>N){for(i=1;i<=N;i++){cin>>point[i].x>>point[i].y;}// end cinfor(i=1;i<=N;i++){for(j=1;j<=N;j++){if(i==j)map[i][j]=0x3f3f3f3f;else{map[i][j]=dis(i,j);}//end if}//end j}// end iprintf("%.2lf\n",Prim());}return 0;}

Kruscal解法

#include<iostream>#include<algorithm>using namespace std;typedef struct tag_point{int x,y,dis;}P;P point[5000];int father[105];int find(int x){//寻找父节点，顺便压缩下路径if(x==father[x])return x;int t=find(father[x]);father[x]=t;return t;}int cmp(const P &a,const P &b){return a.dis<b.dis;}int main(){int n,i,sum,t,x,y,c;//freopen("1.txt","r",stdin);while(scanf("%d",&n)!=EOF&&n){t=(n*(n-1))/2;for(i=0;i<t;i++){scanf("%d%d%d",&point[i].x,&point[i].y,&point[i].dis);}sort(point,point+t,cmp);//从小到大排列for(i=1;i<=n;i++){//初始化father[i]=i;}sum=0;//记住最小生成树的值c=0;for(i=0;i<t && c!=n;i++){x=find(point[i].x);y=find(point[i].y);/if(x!=y){sum+=point[i].dis;c++;father[x]=y;}}printf("%d\n",sum);}return 0;}