随笔 - 87  文章 - 279  trackbacks - 0
<2024年11月>
272829303112
3456789
10111213141516
17181920212223
24252627282930
1234567

潜心看书研究!

常用链接

留言簿(19)

随笔分类(81)

文章分类(89)

相册

ACM OJ

My friends

搜索

  •  

积分与排名

  • 积分 - 214327
  • 排名 - 116

最新评论

阅读排行榜

评论排行榜

BellmanFord实现

The Doors

Time limit: 1 Seconds   Memory limit: 32768K  
Total Submit: 214   Accepted Submit: 63  

You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x = 10, y = 0, and y = 10. The initial and final points of the path are always (0, 5) and (10, 5). There will also be from 0 to 18 vertical walls inside the chamber, each with two doorways. The figure below illustrates such a chamber and also shows the path of minimal length.


Input

The input data for the illustrated chamber would appear as follows.

2
4 2 7 8 9
7 3 4.5 6 7

The first line contains the number of interior walls. Then there is a line for each such wall, containing five real numbers. The first number is the x coordinate of the wall (0 < x < 10), and the remaining four are the y coordinates of the ends of the doorways in that wall. The x coordinates of the walls are in increasing order, and within each line the y coordinates are in increasing order. The input file will contain at least one such set of data. The end of the data comes when the number of walls is -1.


Output

The output file should contain one line of output for each chamber. The line should contain the minimal path length rounded to two decimal places past the decimal point, and always showing the two decimal places past the decimal point. The line should contain no blanks.


Sample Input

1
5 4 6 7 8
2
4 2 7 8 9
7 3 4.5 6 7
-1


Sample Output

10.00
10.06

#include <iostream>
#include 
<cmath>
using namespace std;

const double INF = 2000000000;
const int MAXN = 100;

struct POINT
{
    
double x, y;
}
;
struct EDGE
{
    
int u, v;
}
;

double g[MAXN][MAXN];
EDGE e[MAXN
*MAXN];
int n;
int i, j;
double wX[20];
double pY[20][4];
double x;
POINT p[MAXN];
int pSize;
int eSize;

double Dis(POINT a, POINT b)
{
    
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}


double Cross(double x1, double y1, double x2, double y2, double x3, double y3)
{
    
return (x2-x1)*(y3-y1)-(x3-x1)*(y2-y1);
}


bool IsOk(POINT a, POINT b)
{
    
if (a.x >= b.x) return false;
    
int i, j;
    
bool flag = true;
    i 
= 0;
    
while (wX[i] <= a.x && i < n) {
        i
++;
    }

    
while (wX[i] < b.x && i < n) {
        
if (Cross(a.x, a.y, b.x, b.y, wX[i], 0)
        
*Cross(a.x, a.y, b.x, b.y, wX[i], pY[i][0]) < 0
        
|| Cross(a.x, a.y, b.x, b.y, wX[i], pY[i][1])
        
*Cross(a.x, a.y, b.x, b.y, wX[i], pY[i][2]) < 0
        
|| Cross(a.x, a.y, b.x, b.y, wX[i], pY[i][3])
        
*Cross(a.x, a.y, b.x, b.y, wX[i], 10< 0{
            flag 
= false;
            
goto ou;
        }

        i
++;
    }

    ou:;
    
return flag;
}


double BellmanFord(int beg, int end)
{
    
double d[MAXN];
    
int i, j;
    
for (i=0; i<MAXN; i++{
        d[i] 
= INF;
    }

    d[beg] 
= 0;
    
bool ex = true;
    
for (i=0; i<pSize && ex; i++{
        ex 
= false;
        
for (j=0; j<eSize; j++{
            
if (d[e[j].u] < INF && d[e[j].v] > d[e[j].u] + g[e[j].u][e[j].v]) {
                d[e[j].v] 
= d[e[j].u] + g[e[j].u][e[j].v];
                ex 
= true;
            }

        }

    }

    
return d[end];
}


void Solve()
{
    p[
0].x = 0;
    p[
0].y = 5;
    pSize 
= 1;
    
for (i=0; i<n; i++{
        scanf(
"%lf"&wX[i]);
        
for (j=0; j<4; j++{
            p[pSize].x 
= wX[i];
            scanf(
"%lf"&p[pSize].y);
            pY[i][j] 
= p[pSize].y;
            pSize
++;
        }

    }

    p[pSize].x 
= 10;
    p[pSize].y 
= 5;
    pSize
++;
    
for (i=0; i<pSize; i++{
        
for (j=0; j<pSize; j++{
            g[i][j] 
= INF;
        }

    }

    eSize 
= 0;
    
for (i=0; i<pSize; i++{
        
for (j=i+1; j<pSize; j++{
            
if (IsOk(p[i], p[j])) {
                g[i][j] 
= Dis(p[i], p[j]);
                e[eSize].u 
= i;
                e[eSize].v 
= j;
                eSize
++;
            }

        }

    }

    printf(
"%.2lf\n", BellmanFord(0, pSize-1));
}


int main()
{
    
while (scanf("%d"&n) != EOF)
    
{
        
if (n == -1break;
        Solve();
    }

    
return 0;
}

posted on 2006-10-09 01:05 阅读(481) 评论(0)  编辑 收藏 引用 所属分类: ACM题目

只有注册用户登录后才能发表评论。
网站导航: 博客园   IT新闻   BlogJava   知识库   博问   管理