【魔改】hdu6325 多校赛3G xy排序凸包+llvector模板

凸包算法前的预处理，可以极角排序，也可以按X，Y轴排序，

极角排序需要找到角落里的一个点，Xy轴排序要跑两遍凸包

而本题的要求只要一个上半凸包，并且有X轴从小到大以及字典序限制，完全符合xy排序，直接一个循环就过了

坑点：我输出了整个数组的大小，然后完全无视了，wa了12发orz

学会了cmd 的fc来比较文件，多校赛数据都是20M的orz

#define _CRT_SECURE_NO_WARNINGS
#include<cmath>
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<stack>
#include<vector>
#include<string.h>
#include<queue>
#include<string>
#include<set>
#include<time.h>
using namespace std;
#define rep(i,t,n)  for(int i =(t);i<=(n);++i)
#define per(i,n,t)  for(int i =(n);i>=(t);--i)
#define mmm(a,b) memset(a,b,sizeof(a))
#define eps 1e-6
#define pb push_back
const int maxn = 2e5 + 10;
const int inf = 1e7 + 5;//0x7fffffff;   //无限大
const int MOD = 1013;
typedef long long ll;



struct V {
    ll x, y;
    int rk;
    //V() {}
    void sc() { scanf("%lld%lld", &x, &y); }
    V(ll a=0, ll b=0,int rk=0) : x(a), y(b),rk(rk) { }
    V operator+(V o) { return V(x + o.x, y + o.y); }
    V operator-(V o) { return V(x - o.x, y - o.y); }
    double L() { return sqrt(x * x + y * y); }
    V N() {
        double l = L();
        return V(x / l, y / l);
    }
    V rot(double th) { return V(x * cos(th) - y * sin(th), x * sin(th) + y * cos(th)); }
    V operator*(ll z) { return V(x * z, y * z); }
    ll operator*(V o) { return x * o.x + y * o.y; }
    ll operator|(V o) { return 1ll*x * o.y - 1ll*o.x * y; }
    bool operator ==(V o) { return x == o.x&&y == o.y; }
    void pr() { printf("%lld %lld
", x, y); }
} p[maxn], P[maxn];



int n, top, head;

bool cmp(V A, V B)
{
    if (A.x != B.x)return A.x < B.x;
    if (A.y != B.y)return A.y > B.y;
    return A.rk < B.rk;
}



void smain() {
    int t; cin >> t;
    while (t--) {

        cin >> n;
        rep(i, 1, n)
        { p[i].sc(); p[i].rk = i; }
        top = 0;
        sort(p + 1, p + 1 + n, cmp);
        P[1] = p[1], P[2] = p[2];
        top = 2;
        rep(i, 3, n)  {
            if (p[i] == P[top])continue;
            while ( top > 1 &&( (((p[i] - P[top]) | (P[top - 1] - P[top])) > 0 )||(((p[i] - P[top]) | (P[top - 1] - P[top]))== 0&&P[top].rk>p[i].rk)) )top--;
            P[++top] = p[i];
            
        }
        rep(i, 1, top) {
            i!=top?printf("%d ", P[i].rk): printf("%d
", P[i].rk);
        }

    }
    cin >> n;

}
#define ONLINE_JUDGE
int main() {
    //ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
    FILE *myfile;
    myfile = freopen("C:\Users\acm-14\Desktop\test\G.in", "r", stdin);
    if (myfile == NULL)
        fprintf(stdout, "error on input freopen
");
    FILE *outfile;
    outfile= freopen("C:\Users\acm-14\Desktop\test\out.txt", "w", stdout);
    if (outfile == NULL)
    fprintf(stdout, "error on output freopen
");
    long _begin_time = clock();
#endif
    smain();
#ifndef ONLINE_JUDGE
    long _end_time = clock();
    printf("time = %ld ms.", _end_time - _begin_time);
#endif
    return 0;
}
/*
2
3
0 0
1 0
2 0
4
0 0
3 0
2 0
5 0

5
0 0
1 3
2 1
4 4
5 0

//1 2 4 5
*/