蓝桥题是真的坑  样例上写的是t组数据的格式 但是测试数据根本没有t 都是单组 这不是坑人吗?

其实就是最短路裸题 建图时需要一点计算几何

对于每个矩形只给任意三个顶点 利用向量点乘判断直角点

假设 1和3 2和4 互为对角点 那就满足 x1+x3=x2+x4 且 y1+y3=y2+y4 以此求出最后一个点

 

 

#include <bits/stdc++.h>
using namespace std;
#define eps 1e-8
#define N 10000000000000000000000.0

struct node1
{
    double x;
    double y;
};

struct node2
{
    bool friend operator < (node2 n1,node2 n2)
    {
        return n1.val>n2.val;
    }
    double val;
    int id;
};

priority_queue <node2> que;
node1 point[410];
double e[410][410];
double pre[110],dis[410];
double c;
int book[410];
int n;

double cal(int u,int v)
{
    return sqrt(pow(point[u].x-point[v].x,2)+pow(point[u].y-point[v].y,2));
}

double dijkstra(int ss,int ee)
{
    node2 cur,tem;
    double w;
    int i,u,v;
    while(!que.empty()) que.pop();
    for(i=0;i<n;i++)
    {
        dis[i]=N;
        book[i]=0;
    }
    tem.id=ss,tem.val=0.0;
    que.push(tem);
    dis[ss]=0.0;
    while(!que.empty())
    {
        cur=que.top();
        que.pop();
        u=cur.id;
        if(book[u]) continue;
        book[u]=1;
        for(v=0;v<n;v++)
        {
            if(u!=v&&!book[v])
            {
                w=e[u][v];
                if(dis[v]>dis[u]+w)
                {
                    dis[v]=dis[u]+w;
                    tem.id=v,tem.val=dis[v];
                    que.push(tem);
                }
            }
        }
    }
    return min(min(dis[4*ee],dis[4*ee+1]),min(dis[4*ee+2],dis[4*ee+3]));
}

int main()
{
    double x1,y1,x2,y2,x3,y3,x4,y4,ans;
    int ss,ee,i,j;
    while(scanf("%d%lf%d%d",&n,&c,&ss,&ee)!=EOF)
    {
        ss--,ee--;
        for(i=0;i<n;i++)
        {
            scanf("%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&pre[i]);
            if(fabs((x2-x1)*(x3-x2)+(y2-y1)*(y3-y2))<eps)//2
            {
                x4=x1+x3-x2;
                y4=y1+y3-y2;
            }
            else if(fabs((x3-x2)*(x1-x3)+(y3-y2)*(y1-y3))<eps)//3
            {
                x4=x1+x2-x3;
                y4=y1+y2-y3;
            }
            else//1
            {
                x4=x2+x3-x1;
                y4=y2+y3-y1;
            }
            point[4*i].x=x1,point[4*i].y=y1;
            point[4*i+1].x=x2,point[4*i+1].y=y2;
            point[4*i+2].x=x3,point[4*i+2].y=y3;
            point[4*i+3].x=x4,point[4*i+3].y=y4;
        }

        n*=4;
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                if(i/4==j/4)
                {
                    e[i][j]=cal(i,j)*pre[i/4];
                }
                else
                {
                    e[i][j]=cal(i,j)*c;
                }
            }
        }
        ans=N;
        for(i=0;i<4;i++)
        {
            ans=min(ans,dijkstra(4*ss+i,ee));
        }
        printf("%.1f\n",ans);

    }
    return 0;
}