bzoj千题计划155：bzoj3543: [ONTAK2010]Garden

http://www.lydsy.com/JudgeOnline/problem.php?id=3543

枚举每一个点，作为左下角

然后枚举 相同的x坐标，y坐标 少的那个 作为另一个角

二分判断另外两个角是否存在

#include<map>
#include<vector>
#include<cstdio>
#include<iostream>
#include<algorithm>

using namespace std;

#define N 100001

int x[N],y[N];
int hx[N],hy[N],tx,ty;

map<int,int>fx,fy;

vector<int>sx[N],sy[N];

void read(int &x)
{
    x=0; int f=1; char c=getchar();
    while(!isdigit(c)) { if(c==‘-‘) f=-1; c=getchar(); }
    while(isdigit(c)) { x=x*10+c-‘0‘; c=getchar(); }
    x*=f;
}

int main()
{
    int n;
    read(n);
    int mi=0;
    for(int i=1;i<=n;++i)
    {
        read(x[i]);
        read(y[i]);
        hx[++tx]=x[i];
        hy[++ty]=y[i];
        mi=min(mi,x[i]);
        mi=min(mi,y[i]);
    }
    sort(hx+1,hx+tx+1);
    sort(hy+1,hy+ty+1);
    tx=unique(hx+1,hx+tx+1)-hx-1;
    ty=unique(hy+1,hy+ty+1)-hy-1;
    for(int i=1;i<=n;++i)
    {
        x[i]=lower_bound(hx+1,hx+tx+1,x[i])-hx;
        y[i]=lower_bound(hy+1,hy+ty+1,y[i])-hy;
    }
    mi=max(0,-mi);
    for(int i=1;i<=tx;++i) fx[hx[i]+mi]=i;
    for(int i=1;i<=ty;++i) fy[hy[i]+mi]=i;
    for(int i=1;i<=n;++i)
    {
        sx[x[i]].push_back(y[i]);
        sy[y[i]].push_back(x[i]);
    }
    for(int i=1;i<=tx;++i) sort(sx[i].begin(),sx[i].end());
    for(int i=1;i<=ty;++i) sort(sy[i].begin(),sy[i].end());
    int j,k;
    int sizx,sizy;
    int len;
    int xx,yy;
    int ans=0;
    for(int i=1;i<=n;++i)
    {
        j=upper_bound(sx[x[i]].begin(),sx[x[i]].end(),y[i])-sx[x[i]].begin();
        k=upper_bound(sy[y[i]].begin(),sy[y[i]].end(),x[i])-sy[y[i]].begin();
        sizx=sx[x[i]].size();
        sizy=sy[y[i]].size();
        if(sizx-j<sizy-k)
        {
            for(;j<sizx;++j)
            {
                len=hy[sx[x[i]][j]]-hy[y[i]];
                xx=fx[hx[x[i]]+len+mi];
                if(!xx) continue;
                if(binary_search(sx[xx].begin(),sx[xx].end(),y[i]) && binary_search(sx[xx].begin(),sx[xx].end(),sx[x[i]][j]))
                    ans++;
            }
        }
        else
        {
            for(;k<sizy;++k)
            {
                len=hx[sy[y[i]][k]]-hx[x[i]];
                yy=fy[hy[y[i]]+len+mi];
                if(!yy) continue;
                if(binary_search(sy[yy].begin(),sy[yy].end(),x[i]) && binary_search(sy[yy].begin(),sy[yy].end(),sy[y[i]][k]))
                    ans++;
            }
        }
    }
    cout<<ans;
}