【O(nlgn)判定点在凸包内】Codeforces Round #113 (Div. 2) B

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思路参考了这个bloghttp://hi.baidu.com/aekdycoin/blog/item/7abf85026f0d7e85d43f7cfe.html

复杂度是O(nlgn)

#define N 100005struct node{    double x,y;}a[100005];double cross(node a,node b,node c){//>0，ab在ac顺时针；<0，ab在ac逆时针    return (b.x-a.x)*(c.y-a.y) - (c.x-a.x)*(b.y-a.y);}int n,m;bool chk(node b){    int i,j;    if(cross(a[0],b,a[1]) <= 0)return false;    if(cross(a[0],b,a[n-1]) >= 0)return false;    int l = 0,r = n-1,mid;    int tag = 0;    while(l<=r){        mid = (l+r)>>1;        if(cross(a[0],b,a[mid]) >= 0){            tag = mid;            l = mid+1;        } else r = mid-1;    }    l = tag,r = tag+1;    if(cross(a[l],b,a[r]) <= 0)return false;    return true;}int main(){    cin>>n;    int i,j;    for(i=0;i<n;i++){        cin>>a[i].x>>a[i].y;    }    int ans = 0;    cin>>m;    node b;    for(i=0;i<m;i++){        cin>>b.x>>b.y;        if(chk(b))ans++;    }    if(ans==m)cout<<"YES"<<endl;    else cout<<"NO"<<endl;    return 0;}