zoj 1010 Area【线段相交问题】

链接：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1010

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29328#problem/B

Area

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Time Limit: 2 Seconds      
Memory Limit: 65536 KB      
Special Judge

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Jerry, a middle school student, addicts himself to mathematical research. Maybe the problems he has thought are really too easy to an expert. But as an amateur, especially as a 15-year-old boy, he had done very well. He is so rolling in thinking the mathematical problem that he is easily to try to solve every problem he met in a mathematical way. One day, he found a piece of paper on the desk. His younger sister, Mary, a four-year-old girl, had drawn some lines. But those lines formed a special kind of concave polygon by accident as Fig. 1 shows.

 
Fig. 1 The lines his sister had drawn

"Great!" he thought, "The polygon seems so regular. I had just learned how to calculate the area of triangle, rectangle and circle. I'm sure I can find out how to calculate the area of this figure." And so he did. First of all, he marked the vertexes in the polygon with their coordinates as Fig. 2 shows. And then he found the result--0.75 effortless.

Fig.2 The polygon with the coordinates of vertexes

Of course, he was not satisfied with the solution of such an easy problem. "Mmm, if there's a random polygon on the paper, then how can I calculate the area?" he asked himself. Till then, he hadn't found out the general rules on calculating the area of a random polygon. He clearly knew that the answer to this question is out of his competence. So he asked you, an erudite expert, to offer him help. The kind behavior would be highly appreciated by him.

Input

The input data consists of several figures. The first line of the input for each figure contains a single integer n, the number of vertexes in the figure. (0 <= n <= 1000).

In the following n lines, each contain a pair of real numbers, which describes the coordinates of the vertexes, (xi, yi). The figure in each test case starts from the first vertex to the second one, then from the second to the third, ���� and so on. At last, it closes from the nth vertex to the first one.

The input ends with an empty figure (n = 0). And this figure not be processed.

Output

As shown below, the output of each figure should contain the figure number and a colon followed by the area of the figure or the string "Impossible".

If the figure is a polygon, compute its area (accurate to two fractional digits). According to the input vertexes, if they cannot form a polygon (that is, one line intersects with another which shouldn't be adjoined with it, for example, in a figure with four lines, the first line intersects with the third one), just display "Impossible", indicating the figure can't be a polygon. If the amount of the vertexes is not enough to form a closed polygon, the output message should be "Impossible" either.

Print a blank line between each test cases.

Sample Input

5
0 0
0 1
0.5 0.5
1 1
1 0
4
0 0
0 1
1 0
1 1
0

Output for the Sample Input

Figure 1: 0.75

Figure 2: Impossible

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Source: 
Asia 2001, Shanghai (Mainland China)

算法：计算几何 线段相交问题 面积问题

如果你不了解跨立实验判断线段相交问题，可以参考下我前面的一篇博客：ZOJ 1648 Circuit Board【跨立实验判断线段相交】

题意 + 思路：

给你 N  个点的坐标，点是按照顺序输入的。

每一个点都与它后面的那个点连成一条线段，最后一个点与起点相连。

求组成的多边形的面积。

注意：并不是直接求多边形的面积。

      首先要判断多边形的合法性：无论是凸多边形还是凹多边形都有可能合法。

                              这里的不合法指的是只要存在一条线段与不和它直接首尾相连的任意一条线段相交，

                              就表示面积不可求，输出 impossible

            怎样算是相交？

            首先：直接两条线段两两跨立的当然算是相交了。

            其次：如果相于端点的也算是相交。

1.直接跨立相交，是肯定不行的。

2.相交于端点，也是非法的。

                     

 

关于线段相交于端点：画的详细点就是这样的图形也被认为是不可以求面积的

PS:图形来自 kuangbin 大神

所以这题的关键不是求面积，而是判断线段相交情况。

PS：本来第二次 WA 了马上就可以 AC 的了，判断线段相交的循环中 j < i 写错成了 j < 2

结果自己写这些简单题目的时候心态还是不行，最后就是纠结了一天多，然后又回到了开始的起点。

感谢 GJ 童鞋的提醒，要不然不知道乱改到什么时候去。

分析：

线段相交模板：自己画图理解下，还是推荐直接写自己的模板比较好了。

// = 0 表示,相交于端点也是认定为相交
bool SegmentProperIntersection(Point s1, Point e1, Point s2, Point e2) {
     double c1 = Cross(s2-s1, e1-s1), c2 = Cross(e1-s1, e2-s1),
            c3 = Cross(s1-s2, e2-s2), c4 = Cross(e2-s2, e1-s2);
     if(c1*c2 >= 0 && c3*c4 >= 0) return true;
     return false;
}

然后就是线段的比较判断问题：

注意: 是要与不和当前线段直接相连的每一条线段都比较。

          那么我们每次加入一个点，直接判断新形成的线段与前面所有出现过的不直接相连的比较就可以了

所以应该从第三条线段开始枚举。

因为第一条线段没有线段可以和它比较；

第二条线段前面只有一条线段，而这条线段又是和他直接相连的，所以不用比较。

这样一直到了第 N-1 条线段都可以这么判断下去。

然后注意到最后一条线段是终点和起点相连形成的，所以它不能和第一条线段比较，也不可以和第 n-1 条线段比较。2.

最后就是叉积求面积，不用解释了吧。

//叉积求面积 ，下标从 0 开始
double Area() {
    double area = 0;
    for(int i = 1; i < n-1; i++)
        area += Cross(p[i]-p[0], p[i+1]-p[0]);
    return fabs(area) / 2.0;
}

code:

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<iostream>
using namespace std;

const int maxn = 1000+10;
int n;

struct Point{
    double x,y;
    Point() {}

    Point(double _x, double _y) {
        x = _x;
        y = _y;
    }
    Point operator - (const Point &b) const{
        return Point(x-b.x, y-b.y);
    }
}p[maxn];

double Cross(Point a, Point b) { //叉积
    return a.x*b.y - a.y*b.x;
}

// = 0 表示,相交于端点也是认定为相交
bool SegmentProperIntersection(Point s1, Point e1, Point s2, Point e2) {
     double c1 = Cross(s2-s1, e1-s1), c2 = Cross(e1-s1, e2-s1),
            c3 = Cross(s1-s2, e2-s2), c4 = Cross(e2-s2, e1-s2);
     if(c1*c2 >= 0 && c3*c4 >= 0) return true;
     return false;
}

bool haveCross() {

//从第 三 条线段开始, 一直到第 N-1 条线段
    for(int i = 2; i < n-1; i++) { //检查 Lin(i,i+1) 与前面每一条不直接相连的线段
        for(int j = 1; j < i; j++) {
            if(SegmentProperIntersection(p[i], p[i+1], p[j-1], p[j])) {
                return true;
            }
        }
    }

    //判断最后一条线段p[0]_p[n-1]
    //从第二条线段一直比较到第 n-2 条线段
    for(int i = 1; i < n-2 ; i++) { //p[i]__p[i+1]
        if(SegmentProperIntersection(p[0],p[n-1],p[i],p[i+1])) {
                return true;
            }
    }
    return false;
}

//叉积求面积
double Area() {
    double area = 0;
    for(int i = 1; i < n-1; i++)
        area += Cross(p[i]-p[0], p[i+1]-p[0]);
    return fabs(area) / 2.0;
}

int main()
{
    int test = 0;
    while(scanf("%d", &n) != EOF) {
        if(n == 0) break;

        double x,y;
        for(int i = 0; i < n; i++) {
            scanf("%lf%lf", &x,&y);
            p[i] = Point(x,y);
        }

        if(n < 3) { //如果小于三个点, 肯定不能求面积
            printf("Figure %d: Impossible\n", ++test);
            printf("\n");
            continue;
        }

        if(haveCross()) { //如果有线段相交的情况
            printf("Figure %d: Impossible\n", ++test);
            printf("\n");
            continue;
        }
        else printf("Figure %d: %.2lf", ++test,Area());
        printf("\n");
    }
    return 0;
}

/**
几组全为 impossible 的数据
*/
/*

7
0 0
4 0
2 1
4 3
3 3
3 2
1 3

7
0 0
1 1
3 0
5 1
5 -1
3 0
1 -1

5
0 0
4 0
3 1
2 0
1 1

*/