POJ2826 An Easy Problem?!

　　原题链接：http://poj.org/problem?id=2826

　　这道题要的思路很容易，但把代码写丑也很容易。

　　接到的水的截面体积为0的情况有：

　　1. 两条线段不相交；

　　2. 其中任意一条线段水平；

　　3. 两条线段重合；

　　4. 相交的情况下，最高的端点遮住了次高的端点

　　

　　这道题断断续续搞了一周才搞定，POJ上wa了几个版面。几何题，同样的功能不同的写法是wa与ac的鸿沟，唉～～～

View Code 
 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <stdlib.h>
 4 #include <algorithm>
 5 #include <math.h>
 6 #define INF 1e7
 7 const double eps=1e-8;
 8 
 9 struct point
10 {
11     double x, y;
12     point(){}
13     point(double _x, double _y):x(_x), y(_y){}
14 };
15 
16 double min(double a, double b){return a < b ? a : b;}
17 double max(double a, double b){return a > b ? a : b;}
18 
19 bool inter(point a, point b, point c, point d)
20 {
21     if ( min(a.x, b.x) > max(c.x, d.x) ||
22          min(a.y, b.y) > max(c.y, d.y) ||
23          min(c.x, d.x) > max(a.x, b.x) ||
24          min(c.y, d.y) > max(a.y, b.y) ) return 0;
25     double h, i, j, k;
26     h = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
27     i = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x);
28     j = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x);
29     k = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x);
30     return h * i <= eps && j * k <= eps;
31 }
32  
33 point intersection(point u1,point u2,point v1,point v2)
34 {
35     point ret=u1;
36     double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))
37              /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
38     ret.x+=(u2.x-u1.x)*t;
39     ret.y+=(u2.y-u1.y)*t;
40     return ret;
41 }
42  
43 double dot(point a, point b, point c, point d)
44 {return (a.x - b.x) * (c.y - d.y) - (a.y - b.y) * (c.x - d.x);}
45 void swapPoint(point &a, point &b)
46 {
47     std::swap(a.x, b.x);
48     std::swap(a.y, b.y);
49 }
50  
51 int main()
52 {
53     int n;
54     bool flag;
55     point a, b, c, d;
56     scanf("%d", &n);
57     while(n --)
58     {
59         scanf("%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y);
60         scanf("%lf%lf%lf%lf", &c.x, &c.y, &d.x, &d.y);
61         if((int)a.y == (int)b.y || (int)c.y == (int)d.y || fabs(dot(a, b, c, d)) < eps || !inter(a, b, c, d))
62             puts("0.00");
63         else
64         {
65             point cp = intersection(a, b, c, d);
66             
67             point lowp, upp, tmp;
68             if(a.y < b.y) swapPoint(a, b);
69             if(c.y < d.y) swapPoint(c, d);
70             bool flag;
71             if(a.y < c.y)
72             {
73                 lowp.x = a.x, lowp.y = a.y;
74                 upp.x = c.x, upp.y = c.y;
75                 tmp = intersection(c, d, *new point(-INF, lowp.y), *new point(INF, lowp.y));
76                 flag = inter(a, *new point(a.x, INF), c, d);
77             }
78             else
79             {
80                 lowp.x = c.x, lowp.y = c.y;
81                 upp.x = a.x, upp.y = a.y;
82                 tmp = intersection(a, b, *new point(-INF, lowp.y), *new point(INF, lowp.y));
83                 flag = inter(c, *new point(c.x, INF), a, b);
84             }
85             if(flag)
86                 printf("0.00\n");
87             else
88                 printf("%.2f\n", fabs(dot(lowp, cp, tmp, cp)) / 2.0);
89         }
90     }
91     return 0;
92 }
posted @ 2012-10-31 18:39  芒果布丁  阅读(232)  评论(0)    收藏  举报