HDU 1875 畅通工程再续

来源：互联网发布：苹果手机java编程软件编辑：程序博客网时间：2024/06/06 18:21

还是一道水题只是边需要自己处理由于点很少 n²的跑出来就好了在把对应的各项都改成double型就可以了

#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <queue>#include <map>#include<set>#include<stack>#define bug puts("bugbugbug");using namespace std;typedef long long ll;const int maxm=10005;int f[105];int n,m;struct node{    int x, y;    double v;    bool operator<(const node &b)const    {        if(this->v!=b.v)            return (this->v<b.v);        if(this->x!=b.x)            return (this->x<b.x);        return (this->y<b.y);    }} e[maxm];int find(int x){    return f[x]= f[x]==x?x:find(f[x]);}double Kruscal(){    sort(e, e + m);    int cnt = n;    double ans = 0;    for(int i = 0; i <= n; i++) f[i] = i;    for (int i = 0; i < m; i++)    {        int x = find(e[i].x);        int y = find(e[i].y);        if (x != y)        {            f[x] = y;            cnt--;            ans += e[i].v;            if (cnt == 1) break;        }    }    if(cnt!=1)return -1;    return ans;}double x[1005],y[1005];double D(int i,int j){    return sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));}int main(){   int T_T;   scanf("%d",&T_T);   while(T_T--)   {       m=0;       scanf("%d",&n);       for(int i=0;i<n;i++)        scanf("%lf%lf",&x[i],&y[i]);       for(int i=0;i<n;i++)       {           for(int j=i+1;j<n;j++)           {               double d=D(i,j);               if(d>=10&&d<=1000)               {               e[m].x=i;               e[m].y=j;               e[m++].v=d;               }           }       }       double ans=Kruscal();       if(ans==-1)        puts("oh!");       else        printf("%.1lf\n",ans*100);   }}

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