题目大意：给出n个点的编号和坐标，按逆时针方向连接着n个点，按连接的先后顺序输出每个点的编号。
题目思路：Cross（a,b）表示a,b的叉积，若小于0：a在b的逆时针方向，若大于0a在b的顺时针方向。每次都sort一下，找出在当前点逆时针方向的最远的点。数据很小O(N*N*log（N）)的复杂度0s AC了……
#include<cstdio> #include<cstdlib> #include<cmath> #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<queue> #define INF 0x3f3f3f3f #define MAX 100005using namespace std;int n,pos,ans[MAX];struct node {int id,x,y; }point[MAX];int Dist(int x1,int y1,int x2,int y2) {return sqrt((x1-x2)*(x1-x2)*1.0 + (y1-y2)*(y1-y2)*1.0); }int Cross(int x1,int y1,int x2,int y2,int x3,int y3) {return (x1-x2)*(y1-y3)-(x1-x3)*(y1-y2); }bool cmp(struct node A,struct node B) {int op=Cross(point[pos].x,point[pos].y,A.x,A.y,B.x,B.y);if(op>0)return true;else if(op==0 && Dist(point[pos].x,point[pos].y,A.x,A.y)<Dist(point[pos].x,point[pos].y,B.x,B.y))return true;return false; }int main() {int T,cnt;scanf("%d",&T);while(T--){pos=0;cnt=0;memset(ans,0,sizeof(ans));scanf("%d",&n);scanf("%d%d%d",&point[0].id,&point[0].x,&point[0].y);for(int i=1;i<n;i++){scanf("%d%d%d",&point[i].id,&point[i].x,&point[i].y);if(point[i].y < point[0].y)//找出起点：左下方的点swap(point[0],point[i]);else if(point[i].y==point[0].y && point[i].x < point[0].x)swap(point[0],point[i]);}sort(point,point+n,cmp);ans[cnt++]=point[pos++].id;for(int i=1;i<n;i++){sort(point+i,point+n,cmp);ans[cnt++]=point[pos++].id;}printf("%d ",cnt);for(int i=0;i<cnt;i++)printf("%d%c",ans[i],i==cnt-1?'\n':' ');}return 0; }
