Gym 100883J palprime（二分判断点在凸包里）

题意：判断一堆小点有多少个在任意三个大点构成的三角形里面。

思路：其实就是判断点在不在凸包里面，判断的话可以使用二分来判断，就是判断该点在凸包的哪两个点和起点的连线之间。

代码：

 /** @xigua */
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<vector>
 #include<stack>
 #include<cstring>
 #include<queue>
 #include<set>
 #include<string>
 #include<map>
 #include<climits>
 #define PI acos(-1)
 using namespace std;
 typedef long long ll;
 typedef double db;
 const int maxn = 5e4 + ;
 const int mod = ( << ) - ;
 const int INF = 1e8 + ;
 const ll inf = 1e15 + ;
 const db eps = 1e-;

 struct Node {
     ll x, y, id;
 } po[maxn];
 int n;
 int flag;

 bool cmp(const Node &a, const Node &b) {
     if (a.x == b.x) return a.y < b.y;
     return a.x < b.x;
 }

 ll mul(ll x1, ll x2, ll y1, ll y2) {
     return x1 * y2 - x2 * y1;
 }
 Node S[maxn];
 int top;

 void ch(Node cur) {
     while (top > flag) {
         Node po1 = S[top];
         Node po2 = S[top-];
         ll x1 = po1.x - po2.x, y1 = po1.y - po2.y;
         ll x2 = cur.x - po2.x, y2 = cur.y - po2.y;
         if (mul(x1, x2, y1, y2) >= ) {
             top--;
         }
         else break;
     }
     S[++top] = cur;
 }

 int ok(int mid, ll x, ll y) {
     ll x1 = S[mid].x - S[].x, y1 = S[mid].y - S[].y;
     ll x2 = x - S[].x, y2 = y - S[].y;
     ll tmp = mul(x1, x2, y1, y2);
     if (tmp > ) return ;
     if (tmp == ) return ;
     return ;
 }

 void solve() {
     while (cin >> n) {
         for (int i = ; i <= n; i++) {
             scanf("%I64d%I64d", &po[i].x, &po[i].y);
             po[i].id = i;
         }
         sort(po+, po++n, cmp);
         top = ;
         flag = ;
         for (int i = ; i <= n; i++) {
             ch(po[i]);
         }
         flag = top;
         for (int i = n-; i >= ; i--) {
             ch(po[i]);
         }
         top--;
         int q, ans = ; cin >> q;
         Node tmp[maxn];
         for (int i = ; i <= top; i++)
             tmp[i] = S[i];
         for (int i = ; i <= top; i++)
             S[i] = tmp[top-i+];
         while (q--) {
             ll x, y;
             scanf("%I64d%I64d", &x, &y);
             int l = , r = top; //二分上界和下界
             while (l < r) {
                 int mid = l + r +  >> ;
                 if (ok(mid, x, y))  //在当前这条线之上
                     l = mid;
                 else r = mid - ;
             }
             if (l == top) { //在上界需要特殊判断
                 if (ok(l, x, y) == ) {
                     ll a = x - S[].x, b = y - S[].y;
                     ll dis1 = a * a + b * b;
                     a = S[l].x - S[].x, b = S[l].y - S[].y;
                     ll dis2 = a * a + b * b;
                     if (dis1 <= dis2) ans++;
                 }
                 continue;
             }
             Node xx[];
             xx[] = S[], xx[] = S[l], xx[] = S[l+], xx[] = S[];
             ll are1 = ;
             for (int i = ; i <= ; i++) {
                 ll x1 = xx[i].x - x, y1 = xx[i].y - y;
                 ll x2 = xx[i+].x - x, y2 = xx[i+].y - y;
                 ll tmp = mul(x1, x2, y1, y2);
                 if (tmp < ) tmp = -tmp;
                 are1 += tmp;
             }
             ll x1 = xx[].x - xx[].x, y1 = xx[].y - xx[].y;
             ll x2 = xx[].x - xx[].x, y2 = xx[].y - xx[].y;
             ll are2 = mul(x1, x2, y1, y2);
             //if (are1 < 0) are1 = -are1;
             if (are2 < ) are2 = -are2;
             if (are2 == are1) ans++;    //通过面积来判断，are1代表的是加上该点的
         }
         cout << ans << endl;
     }
 }

 int main() {
     //cin.sync_with_stdio(false);
     //freopen("in.txt", "r", stdin);
     //freopen("isharp.out", "w", stdout);
     int t  = ; //cin >> t;

     while (t--) {
         solve();
     }
     return ;
 }