Codeforces Round #352 (Div. 2) C. Recycling Bottles(枚举)

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

double a[100100],b[100100],t[100100];
double minn[4],sum,chang;
int under[4];

double distance(double x,double y,double x1,double y1){
    return sqrt(pow((x-x1)*1.0,2)+pow((y-y1)*1.0,2)+0.0);
}

int main(){
    //int ax,ay,bx,by,tx,ty,n,tempx,tempy;
    int n;
    double ax,ay,bx,by,tx,ty,tempx,tempy;
    scanf("%lf%lf%lf%lf%lf%lf",&ax,&ay,&bx,&by,&tx,&ty);
    scanf("%d",&n);
    
    sum=0;
    for(int i=1;i<=n;i++){
        scanf("%lf%lf",&tempx,&tempy);    
        t[i] = distance(tx,ty,tempx,tempy);
        sum += 2*t[i];
        a[i] = distance(ax,ay,tempx,tempy) - t[i];
        b[i] = distance(bx,by,tempx,tempy) - t[i];
        //cout<<a[i]<<" "<<b[i]<<endl;
    } 
    
    minn[0] = a[1];under[0] = 1;
    minn[2] = b[1];under[2] = 1;
    
    //找出a[i]数组的两个最小值 
    for(int i=1;i<=n;i++){
        if(a[i]<minn[0]){
            minn[0] = a[i];
            under[0] = i;
        }
    }
    
    for(int i=1;i<=n;i++) if(i!=under[0]) minn[1] = a[i],under[1] = i;
    
    for(int i=1;i<=n;i++){
        if(i!=under[0]&&a[i]<minn[1]){
            minn[1] = a[i];
            under[1] = i;
        }
    }
    
    //找出b[i]数组的两个最小值 
    for(int i=1;i<=n;i++){
        if(b[i]<minn[2]){
            minn[2] = b[i];
            under[2] = i;
        }
    }
    
    for(int i=1;i<=n;i++) if(i!=under[2]) minn[3] = b[i],under[3] = i;
    
    for(int i=1;i<=n;i++){
        if(i!=under[2]&&b[i]<minn[3]){
            minn[3] = b[i];
            under[3] = i;
        }
    }
    
    // 当n=1的时候会有就没有两个最小值了  所以要当成特殊情况?
    if(n==1) chang = min(a[1],b[1]);
    else if(under[0]!=under[2]){
        chang = min(min(minn[0]+minn[2],minn[0]),minn[2]);
    }else{
        chang = min(min(min(minn[0]+minn[3],minn[1]+minn[2]),minn[0]),minn[2]);
    }
    
    printf("%.12lf\n",sum+chang);
    
    return 0;
}