[最短路]luogu P1354 房间最短路问题

题面

https://www.luogu.com.cn/problem/P1354

分析

傻逼题，评级虚高

判断墙点之间是否有墙阻隔，没有连个边，直接跑folyd即可

代码

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const double Inf=21.0;
const int N=22;
struct Wall {
    double x,y[4];
}p[N];
int n;
double g[N][4][N][4];

inline double Dist(double x1,double y1,double x2,double y2) {return sqrt(pow(x1-x2,2)+pow(y1-y2,2));}

inline double Slope(double x1,double y1,double x2,double y2) {return (y2-y1)/(x2-x1);}

inline double Intercept(double x,double y,double k) {return y-k*x;}

inline double Function(double x,double k,double b) {return x*k+b;}

bool Check(int u,int v,double y1,double y2) {
    if (u>=v) return 0;
    double k=Slope(p[u].x,y1,p[v].x,y2),b=Intercept(p[u].x,y1,k);
    for (int i=u+1;i<=v;i++) {
        double f=Function(p[i].x,k,b);
        if (f<p[i].y[0]||p[i].y[1]<f&&f<p[i].y[2]||p[i].y[3]<f) return 0;
    }
    return 1;
}

int main() {
    scanf("%d",&n);n++;
    for (int i=1;i<n;i++) scanf("%lf%lf%lf%lf%lf",&p[i].x,&p[i].y[0],&p[i].y[1],&p[i].y[2],&p[i].y[3]);
    p[0].x=0;p[n].x=10;p[0].y[0]=p[0].y[1]=p[0].y[2]=p[0].y[3]=p[n].y[0]=p[n].y[1]=p[n].y[2]=p[n].y[3]=5;
    for (int i=0;i<=n;i++)
        for (int j=0;j<4;j++)
            for (int k=0;k<=n;k++)
                for (int l=0;l<4;l++)
                    if (Check(i,k,p[i].y[j],p[k].y[l])) g[i][j][k][l]=Dist(p[i].x,p[i].y[j],p[k].x,p[k].y[l]);
                    else g[i][j][k][l]=Inf;
    for (int m=0;m<=n;m++)
        for (int s=0;s<4;s++)
            for (int i=0;i<=n;i++)
                for (int j=0;j<4;j++)
                    for (int k=0;k<=n;k++)
                        for (int l=0;l<4;l++)
                            g[i][j][k][l]=min(g[i][j][k][l],g[i][j][m][s]+g[m][s][k][l]);
    printf("%.2lf",g[0][0][n][0]);
}