poj 1556

几何加floyd 水题一个。浮点数还是看脸。
题意：在10*10的矩形中，给出一些线段，线段分割矩形，求（0，5）到（10，5）的最小距离。数据很小直接暴力。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;
#define count sum
#define eps  0.0000001
int count = 0,n;
struct point{
    double x; //X轴的下标
    double y;  //  Y轴 
};
struct line {   // 直线方程
    double A,B,C;
};
void Coeeficient(line &L,point a,point b) {     // 两点求直线方程
    L.A = a.y - b.y;
    L.B = b.x - a.x;
    L.C = -L.A*a.x-L.B*a.y;
}
point data[200]={};        // 点的统计

double dis(point a,point b) {        // 欧式距离
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double pd(point a,point b) {     // 判断直线与线段是否相交
    if(fabs(a.x - b.x) <0.0000001) return 10000000;
    line L;
    Coeeficient(L,a,b);
    for(int i = 0; i < count; i++) {
        if(data[i].x > a.x  && data[i].x < b.x) {

            double Y = (L.A*data[i].x+L.C)/(-L.B);
            if(i%4 == 0) {
                if(data[i].y > Y)
                  return 100000000;
            }
            else if(i%4 == 1 ) {
                if(Y > data[i].y && data[i+1].y > Y) {return 10000000;}
            }
            else if(i % 4 == 3)
              if(Y > data[i].y) return 1000000;
        }
    }
    return dis(a,b);
}
void solve() {
    double mp[100][100]={};
    for(int i = 0; i < count; i++)
      for(int j = 0; j < count; j++)
        mp[i][j] = 10000000;

    for(int i = 0; i < count; i++)
      for(int j = 0; j < count; j++) {
          double dis = pd(data[i],data[j]);
          mp[i][j] = dis;
      }

    for(int k = 0; k < count; k++)
      for(int i = 0; i < count; i++)
        for(int j = 0; j < count; j++)
          mp[i][j] = min(mp[i][j],mp[i][k]+mp[k][j]);
    printf("%.2lf\n",mp[count-2][count-1]);
}
int main() {
    while(scanf("%d",&n),n != -1) {
        count = 0;
        double x;
        for(int i = 0; i < n; i++) {
            scanf(" %lf",&x);
            data[count].x = data[count+1].x = data[count+2].x = data[count+3].x = x;
            for(int j = 0; j < 4; j++)
              scanf(" %lf",&data[count+j].y);
            count += 4;
        }
        data[count].x = 0; data[count++].y = 5;
        data[count].x = 10; data[count++].y = 5;
        solve();
    }
    return 0;
}