Poj1228 Grandpa's Estate

来源:互联网 发布:网络证券交易平台 编辑:程序博客网 时间:2024/06/02 17:33

给出了一些凸包上的点,问该凸包是否为稳定凸包。
稳定凸包即指在不删掉当前凸包上的点的情况下,无法通过加点来得到更大的凸包。这样的凸包每条边上除两端点外都一定还有点。
求出凸包顶点,判断凸包每条边上是否还有点即可。
既然那么多人都写Graham,那我就写写分治来愉悦一下吧。

#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<cmath>using namespace std;const int maxn = 1005;#define EPS 1e-6#define INF 1e9struct point {    double x,y;    point(){}    point(double _x,double _y):x(_x),y(_y){}    point operator - (const point &a) {return point(x-a.x,y-a.y);}    bool operator != (const point &a) {return x!=a.x || y!=a.y;}    bool operator == (const point &a) {return x==a.x && y==a.y;}};double multi(point a,point b) {return a.x*b.y - a.y*b.x;}double dis(point a,point b) {point c=a-b;return sqrt(1.0*c.x*c.x+1.0*c.y*c.y);}bool cmp(point a,point b) {return a.x < b.x || (a.x==b.x && a.y < b.y);}point P[maxn],hull[maxn];int N,cnt=0;double s[maxn];void quickhull(int L,int R,point a,point b) {    int x = L,i=L-1,j=R+1;    for(int k = L; k <= R; k++) if( s[k]-s[x] > EPS || ( fabs(s[x]-s[k])<EPS && cmp(P[x],P[k]))) x=k;    point y = P[x];    for(int k = L; k <= R; k++) {        s[++i] = multi(a-P[k],y-P[k]);        if( s[i] > EPS) swap(P[i],P[k]); else i--;    }    for(int k = R; k >= L; k--) {        s[--j] = multi(y-P[k],b-P[k]);        if( s[j] > EPS) swap(P[j],P[k]); else j++;    }    if( L <= i ) quickhull(L,i,a,y);    hull[++cnt] = y;    if( j <= R ) quickhull(j,R,y,b);}bool judge(point a,point b,point k) {    return fabs(multi(k-a,b-a)) < EPS;}int main() {    int T;    for(scanf("%d",&T);T;T--) {        scanf("%d",&N);        memset(s,0,sizeof s);        P[0] = point(INF,INF); int x = 0;        for(int i = 1; i <= N; i++) {            scanf("%lf%lf",&P[i].x,&P[i].y);            if( cmp(P[i],P[x]) ) x=i;        }        if(N< 5) {puts("NO"); continue;}        swap(P[1],P[x]);        cnt=0;        hull[++cnt] = P[1];        quickhull(2,N,P[1],P[1]);        hull[++cnt] = P[1];        bool fg = 1;        for(int i = 2; i <= cnt; i++) {            point a = hull[i-1];            point b = hull[i];            bool found=0;            for(int j = 1; j <= N; j++){                if(P[j] == a || P[j] == b) continue;                if(judge(a,b,P[j])) {found=1;break;}            }            if( !found ){ fg = 0; break;}        }        puts( fg ? "YES" : "NO");    }}
0 0
原创粉丝点击