【NOIP2001】【Luogu1027】Car的旅行路线

problem

solution

codes

//1.计算几何求第四点坐标, 方法很多
//2.虚点，到A城市的四个机场边权都为0
//3.SPFA跑最短路
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<vector>
#include<queue>
using namespace std;

//Timu
double n, m, a, b;
double x[500], y[500], t[110];

//Graph
struct Edge{
    int to;  double w;
    Edge(int x, double y):to(x),w(y){}
};
vector<Edge>G[500];
void insert(int u,int v,double w){
    G[u].push_back(Edge(v,w));
}
double distant(double nx,double ny,double mx,double my){
    return sqrt((nx-mx)*(nx-mx)+(ny-my)*(ny-my));
}

//spfa
queue<int>q;
int vis[500];
double dis[500];
void spfa(){
    while(!q.empty()){
        int u = q.front();  q.pop();  vis[u] = 0;
        for(int i = 0; i < G[u].size(); i++){
            int v = G[u][i].to;
            if(dis[u]+G[u][i].w<dis[v]){
                dis[v] = dis[u]+G[u][i].w;
                if(!vis[v]){
                    q.push(v);
                    vis[v] = 1;
                }
            }
        }
    }
}

int main(){
    int T;  cin>>T;
    while(T--){
        //1.初始化
        memset(x,0,sizeof x);
        memset(y,0,sizeof y);
        memset(t,0,sizeof t);
        memset(vis,0,sizeof vis);
        //2.datein
        cin>>n>>m>>a>>b;
        for(int i = 0; i < n; i++){
            for(int j = 1; j < 4; j++)
                cin>>x[i*4+j]>>y[i*4+j];
            cin>>t[i+1];
            //point4, 
            double l1=distant(x[i*4+1],y[i*4+1],x[i*4+2],y[i*4+2]);
            double l2=distant(x[i*4+1],y[i*4+1],x[i*4+3],y[i*4+3]);
            double l3=distant(x[i*4+2],y[i*4+2],x[i*4+3],y[i*4+3]);
            double l=max(l1,max(l2,l3));//三条里最常的就是对角线,然后中点坐标得到第4点
            if(l1 == l)x[i*4+4]=(x[i*4+1]+x[i*4+2])-x[i*4+3], y[i*4+4]=(y[i*4+1]+y[i*4+2])-y[i*4+3];
            if(l2 == l)x[i*4+4]=(x[i*4+1]+x[i*4+3])-x[i*4+2], y[i*4+4]=(y[i*4+1]+y[i*4+3])-y[i*4+2];
            if(l3 == l)x[i*4+4]=(x[i*4+2]+x[i*4+3])-x[i*4+1], y[i*4+4]=(y[i*4+2]+y[i*4+3])-y[i*4+1];
        }
        //3.预处理,建图(把所有机场连起来就好啦啦啦~)
        n *= 4;
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                if(i == j)continue;
                int c1 = (i-1)/4+1, c2 = (j-1)/4+1;
                if(c1 == c2){
                    double w = distant(x[i],y[i],x[j],y[j])*t[c1];
                    insert(i,j,w);
                    insert(j,i,w);
                }else{
                    double w = distant(x[i],y[i],x[j],y[j])*m;
                    insert(i,j,w);
                    insert(j,i,w);
                }
            }
        }
        //4.run
        for(int i = 1; i <= n; i++)dis[i] = 1e9+1;
        for(int i = (a-1)*4+1; i <= a*4; i++){
            q.push(i);
            dis[i] = 0;
            vis[i] = 1;
        }
        spfa();
        //5.dateout
        double ans = 1e9;
        for(int i = (b-1)*4+1; i <= 4*b; i++)ans = min(ans, dis[i]);
        printf("%.1lf
",ans);
    }
    return 0;
}