凸包应用

http://acm.hnu.cn/online/?action=problem&type=show&id=13343

Around the Track

Time Limit: 1000ms, Special Time Limit:2500ms,Memory Limit:32768KB

Total submit users: 0, Accepted users:0

Problem 13343 : No special judgement

Problem description

In order to compare race tracks, we wish to compute their lengths. A racetrack is strictly two-dimensional (no elevation). It is described by two simple polygons, where one is completely contained inside the other. The track is the region between these two polygons. We define the length of the track as the absolute minimum distance that one needs to travel in order to complete a lap. This could involve traveling on the very edge of the track and arbitrarily sharp cornering (see Figure A.1).

Input

The input consists of: • one line with one integer n (3 ≤ n ≤ 50), the number of vertices of the inner polygon; • n lines, the ith of which contains two integers xiand yi(−5000 ≤ xi,yi≤ 5000): the coordinates of the ith vertex of the inner polygon; • one line with one integer m (3 ≤ m ≤ 50), the number of vertices of the outer polygon; • m lines, the ith of which contains two integers xiand yi(−5000 ≤ xi,yi≤ 5000): the coordinates of the ith vertex of the outer polygon. For both polygons, the vertices are given in counterclockwise order. The borders of the two polygons do not intersect or touch each other.

Output

Output one line with one floating point number: the length of the race track. Your answer should have an absolute or relative error of at most 10^(−6).

Sample Input

3 1 1 2 1 1 2 3 0 0 4 0 0 4 5 1 1 5 1 5 5 3 3 1 5 4 0 0 6 0 6 6 0 6 5 1 1 5 1 5 5 3 3 1 5 5 0 0 6 0 6 6 3 4 0 6

Sample Output

3.41421356237309 16 16.4721359549996

思路：

1，求出外圈凸包，将不在外圈凸包中的外圈上的点加入集合B.

2，求出内圈凸包.

3，逐个判断B中的点是否在内圈凸包内.

4，排序，求和。

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

int vis[55];
bool mark[55];
typedef struct Point{
    int x, y;
    bool operator < (const Point& other) const{
        if (y != other.y)
            return y < other.y;
        return x < other.x;
    }
}Point;
Point p[55],q[105];
Point ans[105];
int Cross(Point a,Point b,Point c){
    return (b.x - a.x)*(c.y - a.y) - (c.x - a.x)*(b.y - a.y);
}
bool cmp(Point a,Point b){
    return Cross(p[0], a, b)>0;
}
bool _cmp(Point a,Point b){
    return Cross(q[0], a, b)>0;
}
bool __cmp(Point a,Point b){
    return Cross(ans[0], a, b)>0;
}
int main(){
    int N;
    int m,l, i, j;

    //scanf("%d", &N);
    while (scanf("%d", &l)!=EOF){
        //ÊäÈëÄÚȦ  ----------q
        for(i=0;i<l;i++)   scanf("%d%d",&q[i].x,&q[i].y);
       //ÊäÈëÍâȦ   ----------p
        scanf("%d",&m);
        for (i = 0; i < m; i++)  scanf("%d%d",&p[i].x,&p[i].y);
        sort(p, p + m);
        sort(p, p + m, cmp);
        vis[0] = 0;vis[1] = 1;
        ans[0] = p[0]; ans[1] = p[1];
        for (j = i = 2; i < m; i++){
            while (j>1&&Cross(ans[j - 1], ans[j - 2], p[i])>0)
                    j--;
            vis[j]=i;
            ans[j++] = p[i];
        }
        memset(mark,0,sizeof(mark));
        for(i=0;i<j;i++) mark[vis[i]]=1;

        sort(q, q + l);  sort(q, q + l, _cmp);
        ans[0] = q[0];  ans[1] = q[1];

        for (j = i = 2; i < l; i++){
            while (j>1&&Cross(ans[j - 1], ans[j - 2], q[i])>0)
                    j--;
            //vis[j]=i;
            ans[j] = q[i];
            j++;
        } //for
        sort(ans, ans + j);
        sort(ans, ans + j, __cmp);
        int k,t=j;
        int pos[55];
        for(i=0;i<m;i++) if(!mark[i]){
            for(k=0;k<j;k++)
                if(Cross(ans[k],ans[k-1],p[i])<0)
                    {ans[t++]=p[i];}
        }
        sort(ans, ans + t);
        sort(ans, ans + t, __cmp);
        double sum=0;
        for (i = 0; i < t; i++) {
            sum+=sqrt((ans[i].x-ans[(i+1)%t].x)*(ans[i].x-ans[(i+1)%t].x)+(ans[i].y-ans[(i+1)%t].y)*(ans[i].y-ans[(i+1)%t].y));
        }
        printf("%lf\n",sum);
    }
    return 0;
}