题目来源:

http://acm.fzu.edu.cn/problem.php?pid=2160

 

代码如下:

using namespace std ;
typedef long long LL ;
const int Max_N = 100005;
struct Point{
    LL x,y;
};
Point data[Max_N];
int ans[Max_N], Next[Max_N];// Next[i]表示i节点的下一个可连接节点
int n;

// p0p1 X p0p2 左转为>0 返回0表示 不可通过
int judge(Point p1, Point p2, Point p0){
    LL c= (p1.x - p0.x)*(p2.y-p0.y) - (p1.y - p0.y)*(p2.x-p0.x);
    if(c>=0) return 0;
    else return 1;
}

void solve(){
    ans[n]=0;
    ans[n-1]=1;
    Next[n-1]=n;
    int u;
    for(int i=n-2; i>=1; i--){
        u=i+1;
        while(u!=n && judge( data[Next[u]], data[u],data[i] ))
            u=Next[u];
        Next[i]=u;
        ans[i] = ans[u] +1;
    }
}
int main(){
    int t,k=1;
    scanf("%d",&t);
    while(t--){
       // cin>>n;
        scanf("%d",&n);
        for(int i=1; i<= n; i++)
            scanf("%I64d%I64d",&data[i].x, &data[i].y);
        solve();
        printf("Case#%d:",k++);
        for(int i=1; i<=n; i++)
        printf(" %d",ans[i]);
        printf("\n");
    }
    return 0;
}