Two-dimensional coordinate system of the dot product, cross product, polygon area

Two-dimensional coordinate system of the dot product, cross product, polygon area

The dot product is defined

With vector \ (\ vec {a} \ ) and \ (VEC {B} \ \) , the definition of the dot product \ (\ vec {a} \ cdot \ vec {b} \) is a real number , a value of the vector \ (\ vec {a} \) in the vector \ (\ vec {b} \ ) on the projected length of the vector is multiplied by \ (\ vec {b} \ ) of the mold

The dot product of nature

Commutative, associative

Dot product calculation

Direct calculation

\ [| \ Begin {and} \ ln \ begin {b} | = | \ Begin {a} | \ Times | \ begin {b} | \ times cos <\ begin {and} \ \ begin {b}> \]

Calculating coordinates

On the coordinate system set \ (\ vec {a} = (x_ {1}, \ y_ {1}), \ \ vec {b} = (x_ {2}, \ y_ {2}) \)

Then \ (\ vec {a} \ cdot \ vec {b} = x_ {1} x_ {2} + y_ {1} y_ {2} \)

The definition of cross product

With vector \ (\ vec {a} \ ) and \ (\ VEC {B} \) , define the cross product \ (\ vec {a} \ times \ vec {b} \) is the new vector .
Which mold length \ (\ vec {a} \ ) and \ (\ vec {b} \ ) enclosed by a parallelogram area.
Direction \ (\ vec {a} \ ) and \ (\ vec {b} \ ) are vertical.
If \ (\ vec {b} \ ) in \ (\ vec {a} \ ) in the counterclockwise direction, i.e., the left-handed presented, then the \ (\ vec {a} \ times \ vec {b} \) is positive; conversely It is negative.

The cross product of nature

Is the cross product mold parallelogram area, the seeking of a triangle in the area of ​​the rectangular coordinate system can be used to calculate the cross product, i.e.,

\ [S_ {OAB} = \ frac {1} {2} | \ begin {a} \ times \ begin {b} | \]

Cross product is not commutative, since the direction change, i.e.
\ [\ vec {a} \ times \ vec {b} = - \ vec {b} \ times \ vec {a} \]

Calculating a two-dimensional cross product mold

Direct calculation

\ [| \ Begin {and} \ times \ begin {b} | = | \ Begin {a} | \ Times | \ begin {b} | \ times sin <\ begin {and} \ \ begin {b}> \]

Calculating coordinates

On the coordinate system set \ (\ vec {a} = (x_ {1}, \ y_ {1}), \ \ vec {b} = (x_ {2}, \ y_ {2}) \)

After deduced \ (| \ vec {a} \ times \ vec {b} | = | x_ {1} y_ {2} - x_ {2} y_ {1} | \)

It will have to remove the absolute value of the directed area

If the value is positive, indicating \ (\ vec {b} \ ) in \ (\ vec {a} \ ) counterclockwise; if the value is negative, explaining \ (\ vec {b} \ ) in \ (\ VEC {a} \) in the clockwise direction.

Calculated polygon area

The basic idea is to divide the polygon into a plurality of triangles, and the area of ​​the cross-product calculation. The idea for the establishment of concave and convex polygons.

Note polygon vertices are arranged counterclockwise \ (P_ {0}, \ P_ {1}, \ .., \ P_ {n - 1} \)

For ease of calculation, select the coordinate origin \ (O (0, \ 0 ) \) as the source point, individually calculating \ (\ VEC {OP_ {I}} \ Times \ VEC {OP_ {I +. 1}} \) , total sum can be.

The formula is

\[ \sum_{i = 0}^{n - 1}(x_{i}y_{i + 1} - x_{i + 1}y_{i}) \]

When \ (i = n - 1 \ ) , the next coordinates to return to \ ((X_ {0}, Y_ {0}) \) , where Japanese sentence can also be a modulo operation.

Complexity \ (O_ {n} \)

Since the area of ​​the cross product having direction, the excess area will be offset, so this algorithm is correct, strict proof is omitted here.

\(code\)

int n;
struct Cor
{
    int x, y;
}cor[105];

inline int read()
{
    char c = getchar();
    int ans = 0, f = 1;
    while(!isdigit(c)) {if(c == '-') f = -1; c = getchar();}
    while(isdigit(c)) {ans = ans * 10 + c - '0'; c = getchar();}
    return ans * f;
}

int main()
{
    while(scanf("%d", &n) && n) {
        for(int i = 0; i < n; ++i)
            cor[i].x = read(), cor[i].y = read();
        double ans = 0.0;
        for(int i = 0; i < n; ++i){
            ans += 0.5 * (cor[i % n].x * cor[(i + 1) % n].y - cor[i % n].y * cor[(i + 1) % n].x);
        }
        printf("%.1f\n", ans);
    }
    return 0;
}