Kyeremal-poj1113-Wall-convex hull

Kyeremal-poj1113-Wall-convex hull

Poj1113-Wall

 

Wall Time Limit:1000 MS   Memory Limit:10000 K Total Submissions:31394   Accepted:10610 Description Once upon a time there was a greedy King who ordered his chief implements ECT to build a wall around the King's castle. the King was so greedy, that he wocould not listen to his own ect's proposals to build a beautiful brick wall with a perfect shape and nice tall towers. instead, he ordered to build the wall around the whole castle using the least amount of stone and labor, but demanded that the wall shocould not come closer to the castle than a certain distance. if the King finds that the role ECT has used more resources to build the wall than it was absolutely necessary to satisfy those requirements, then the role ect will loose his head. moreover, he demanded effecect to introduce at once a plan of the wall listing the exact amount of resources that are needed to build the wall. Your task is to help poor revoke ECT to save his head, by writing a program that will find the minimum possible length of the wall that he cocould build the castle to satisfy King's requirements. The task is somewhat simplified by the fact, that the King's castle has a polygonal shape and is situated on a flat ground. the specified ECT has already established a Cartesian coordinate system and has precisely measured the coordinates of all castle's vertices in feet. Input The first line of the input file contains two integer numbers N and L separated by a space. N (3 <= N <= 1000) is the number of vertices in the King's castle, and L (1 <= L <= 1000) is the minimal number of feet that King allows for the wall to come close to the castle. Next N lines describe coordinates of castle's vertices in a clockwise order. each line contains two integer numbers Xi and Yi separated by a space (-10000 <= Xi, Yi <= 10000) that represent the coordinates of ith vertex. all vertices are different and the sides of the castle do not intersect anywhere except T for vertices. Output Write to the output file the single number that represents the minimal possible length of the wall in feet that cocould be built around the castle to satisfy King's requirements. you must present the integer number of feet to the King, because the floating numbers are not supported Ted yet. however, you must round the result in such a way, that it is accurate to 8 inches (1 foot is equal to 12 inches), since the King will not tolerate larger error in the estimates. Sample Input 9 100200 400300 400300 300400 300400 400500 400500 200350 200200 200 Sample Output 1628 Hint The result is rounded to the nearest integer. Source Northeastern Europe 2001.

 

Given N points, find the smallest graph so that all points are within the convex polygon, and the distance between any point and the convex polygon is not less than L, and find the smallest perimeter of the graph.

 

First think of a convex bag.

But there is a distance not less than L limit, so consider each point as a circle with a radius of L.

The final graph must contain all N circles.

The final graph includes a straight line and an arc.

The straight line is tangent to the circle, and the Sigma length of the corresponding line is equal to the convex circumference,

Next, consider the Σ arc length. Each section of the arc corresponds to a turning point.

Consider a point of I on the circle and return to the original position after each arc.

That is, Σ arc lengths = the circumference of the circle with the radius of L.

Then the final answer = convex circumference + 2 × π × R.

 

Code:

 

#include 
 
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   #include 
   
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     #include 
     
      #include using namespace std;#define rep(i, l, r) for (int i = l; i <= r; i++)#define REP(i, l, r) for (int i = l; i >= r; i--)#define PI (3.14159265358979323846264)#define INF 19971228#define MAXN 1010int n, L, stack[MAXN], top;struct point {int x, y;} a[MAXN], p0;point operator -(point a, point b) {a.x = a.x - b.x, a.y = a.y - b.y; return a;}int operator ^(point a, point b) {return a.x*b.y - a.y*b.x;}int operator *(point a, point b) {a = a - p0, b = b - p0; return a^b;}inline int sqr(int x) {return x*x;}inline double dist(point a, point b) {return sqrt(sqr(a.y-b.y) + sqr(a.x-b.x));}inline bool cmp(point a, point b) {    if (a*b > 0) return 1;    else if (a*b == 0 && dist(a, p0) < dist(b, p0)) return 1;    return 0;}inline bool left(point a, point b, point c) {    point v1, v2;    v1.x = b.x - a.x, v1.y = b.y - a.y;    v2.x = c.x - b.x, v2.y = c.y - b.y;    return (v1^v2) > 0;}inline void graham() {    a[++n] = a[1];    top = 0;    stack[++top] = 1;    stack[++top] = 2;    stack[++top] = 3;    rep(i, 4, n) {while (!left(a[stack[top-1]], a[stack[top]], a[i])) top--;stack[++top] = i;    }}int main() {    cin >> n >> L;    int minx = INF, miny = INF, k;    rep(i, 1, n) {scanf("%d%d", &a[i].x, &a[i].y);if (a[i].x < minx || (a[i].x == minx && a[i].y < miny)) minx = a[i].x, miny = a[i].y, k = i;    }    p0.x = minx, p0.y = miny;    sort(a+1, a+1+n, cmp);    graham();    double ans = 0;    rep(i, 2, top) ans += dist(a[stack[i-1]], a[stack[i]]);    ans += 2 * PI * L;    printf("%d\n", int(ans + 0.5));    return 0;}