zoj 1203 Swordfish_最小生成树

来源：互联网发布：mysql 合计编辑：程序博客网时间：2024/06/01 07:26

首先我觉得这题格式超级坑，另外就是要注意权值是浮点型！！！

#include <iostream>#include <algorithm>#include <iomanip>#include <cstring>#include <cmath>using namespace std;int n;int Father[5005];struct node{    int u,v;    double w;};node Edge[5005];bool cmp(node a,node b){    return a.w<b.w;}double dist(double x1,double y1,double x2,double y2) //求（x1,y1）和(x2,y2)之间的距离{    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}void MakeSet()  //下面三个函数是并查集  判断时候会生成环{    for(int i=1; i<=n; i++)        Father[i]=i;}int Find(int x){    while(x!=Father[x])        x=Father[x];    return x;}void Union(int x,int y){    int a=Find(x);    int b=Find(y);    if(a!=b)        Father[a]=b;}int main(){    int t,i,j,k,x,y,z,p=1,d=0;    double a[5005],b[5005];    while(cin>>t,t)    {        x=0,y=0,z=0;        double sum=0;        n=t*(t-1)/2;        MakeSet();   //很容易忘记写！！        for(i=1; i<=t; i++)            cin>>a[i]>>b[i];        for(i=1; i<=t; i++)  //将任意两点间的距离和顶点放入结构体里面            for(j=i+1; j<=t; j++)            {                Edge[x++].u=i;                Edge[y++].v=j;                Edge[z++].w=dist(a[i],b[i],a[j],b[j]);            }        sort(Edge,Edge+n,cmp);        for(k=0; k<n; k++)            if(Find(Edge[k].u)!=Find(Edge[k].v))            {                sum=sum+Edge[k].w;                Union(Edge[k].u,Edge[k].v);            }        if(d>0)   //控制格式（超级坑！！！）            cout<<endl;        cout<<"Case #"<<p++<<":"<<endl;        cout<<"The minimal distance is: ";        cout<<setiosflags(ios::fixed)<<setprecision(2)<<sum<<endl;        d++;    }    return 0;}

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