ZOJ3234 Protect the King

题什么的都不用说了吧，直接切正题：给N个点，求能画出的不相交矩形数量最多。然后就直接一个一个的求凸包，知道剩下三个以下的点。就是答案了。

我用的水平序，搞一下就好了。

附代码：

View Code

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define MaxN 2010
struct atp
{
    int x,y;
    }p[MaxN];
int s[MaxN],a[MaxN];
bool v[MaxN],use[MaxN];
int n,tot,ans;
bool cmp(atp a,atp b)
{
    if (a.y<b.y) return true; else
    if (a.y>b.y) return false; else
    if (a.x<=b.x) return true; else return false;
    }

long long cross(int i,int j,int k)
{
     long long x1=p[j].x-p[i].x,
                      y1=p[j].y-p[i].y,
                      x2=p[k].x-p[i].x,
                      y2=p[k].y-p[i].y;
     return x1*y2-x2*y1;
}

void work()
{
    memset(v,0,sizeof(v));
    int i=1,top=2,cnt=0;
    while (use[i]) i++; s[1]=i++;
    while (use[i]) i++; s[2]=i++;
    for (;i<=n;i++) if (!use[i])
    {
        while (top>1 && cross(s[top-1],s[top],i)<0) top--;
        s[++top]=i;
        }
    for (i=1;i<=top;i++)
    {
        a[++cnt]=s[i];
        v[s[i]]=true;
        }
    i=n,top=2;
    while (use[i]) i--;s[1]=i--;
    while (use[i]) i--;s[2]=i--;
    for (;i>=1;i--) if (!use[i])
    {
        while (top>1&&cross(s[top-1],s[top],i)<0) top--;
        s[++top]=i;
        }
        
    for (i=1;i<=top;i++)
        if (!v[s[i]]) a[++cnt]=s[i];
            
    for (i=1;i<=cnt;i++) use[a[i]]=true;
    tot-=cnt;
    ans++;    
        }
    
int main()
{
    freopen("in","r",stdin);
    freopen("out","w",stdout);
    while (scanf("%d",&n)!=EOF)
    {
        tot=n;ans=0;
        for (int i=1;i<=n;i++)
            scanf("%d%d",&p[i].x,&p[i].y);
        sort(p+1,p+n+1,cmp);
        memset(use,false,sizeof(use));
        while (tot>2) work();
        printf("%d\n",ans);
        }
    return 0;
    }