poj1556

题目大意：在10*10的空间里，中间有几堵平行于竖直方向的墙，每个墙由两扇门，求从（0，5）-->(10,5)

的最短路径

思路：把门边沿的点两点建立一条边（中间无墙挡住），和起点，终点同样建边。。最后就成最短路。。

         网上大部分Dijkstra，其实数据很小，floyed足以搞定。。

 PS：这道题数据很弱，但是弱菜下标写错了。。调了非常久。。。还有看到网上有的写了接近300行代码吓尿了。。

写的很搓的代码：

/*
  Time:2013-03-25 23:56:19
  State： Accepted 
*/

#include<iostream>
#include<cstring>
#include<string>
#include<cstdlib>
#include<cstdio>
#include<fstream>
#include<cmath>
#include<algorithm>
#define INF 1000000
#define ee 1e-8
using namespace std;
int n, tot; 
double a[200][6] , map[202][202] ;
int pos[200][6],  p1,  p2 , st , en;


void init(){
     memset(a, 0, sizeof(a));
     tot = 0;
     for (int i = 0;  i<= 200; ++i)
         for (int j = 0; j <= 200; ++j)
            map[i][j] = INF;
     for (int i = 1;  i <= n; ++i){
         scanf("%lf%lf%lf%lf%lf",&a[i][0], &a[i][1] , 
                     &a[i][2] , &a[i][3], &a[i][4] );
         for (int j = 1; j <= 4; ++j) pos[i][j] = i + j * n - n;
     }
}

double work_x(double x2, double y2, double x1 , double y1){
      return x1*y2 - x2*y1;
}

void work(double x1 , double y1, double x2 , double y2){
     double x , y3 , y4;
     bool flag;
     for (int i = st; i <= en ; ++i){
         flag = 0;
         x = a[i][0]; 
         y3 = a[i][1];
         y4 = a[i][2];
         if (work_x(x1 - x , y1 - y3,  0, y4-y3)*work_x(x2 -x , y2 - y3, 0, y4-y3) < 0
            && work_x(x-x1 , y3-y1, x2 - x1, y2-y1)*work_x(x-x1 , y4-y1, x2-x1, y2-y1) < 0) flag = 1;
         y3 = a[i][3];
         y4 = a[i][4];     
         if (work_x(x1 -x , y1 - y3,  0, y4-y3)*work_x(x2 -x , y2 - y3, 0, y4-y3) < 0
            && work_x(x-x1 , y3-y1, x2 - x1, y2-y1)*work_x(x-x1 , y4-y1, x2-x1, y2-y1) < 0) flag = 1;
         if (!flag) return;
      
     }
    map[p2][p1] = map[p1][p2] = sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
}

void solve(){
     tot = n * 4 + 1;
     for (int i = 1; i <= n; ++i)
      for (int k = 1; k <= 4; ++k){
      /*  for (int k1 = k + 1; k1 <= 4; ++k1)
                  map[pos[i][k1]][pos[i][k]] =
                  map[pos[i][k]][pos[i][k1]] = a[i][k1] - a[i][k]; */ 
        p1 = pos[i][k];
        p2 = tot;
        en = n;
        st = i + 1;
        work(a[i][0],a[i][k],10,5);
        p1 = 0;
        p2 = pos[i][k];
        en = i - 1;
        st = 1;
        work(0,5,a[i][0],a[i][k]);
        for (int j = 1; j < i; ++j)
          for (int l = 1; l <= 4 ; ++l){
               st = j + 1;
               p1 = pos[j][l];
               work(a[j][0], a[j][l], a[i][0], a[i][k]);
          }
        
      }   
      p1 =0;
      p2 = tot;
      en = n;
      st = 1;
      work(0,5,10,5);
    /*  for (int i = 0; i <= tot; ++i)
        for (int j =i + 1; j <=tot; ++j)
          if (map[i][j]!=INF) printf("map[%d][%d] = %lf\n",i , j ,map[i][j]);*/
      for (int k = 0; k <= tot; ++k)
        for (int i = 0; i <= tot; ++i)
           for (int j = 0; j <= tot; ++j){
                if (i == j || j == k || k == i) continue;
                map[i][j] = min(map[i][j], map[i][k] + map[k][j]);
           }
      printf("%0.2f\n",map[0][tot]);
     
             
}

int main(){
    freopen("poj1556.in","r",stdin);
    freopen("poj1556.out","w",stdout);
    while (scanf("%d", &n) != EOF && n != -1){
         init();
         solve();    
    }
    fclose(stdin); fclose(stdout);   
}