ZOJ 1426 Counting Rectangles

枚举两条竖线之间同时与这两条竖线相交的横线的个数cnt，这两条竖线之间能组成矩形的个数为：从cnt中任选两条横线的方案个数。

将每两条竖线之间能组成矩形的个数加起来就是结果。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>

using namespace std;

const int MAXN = 110;

struct MyLine
{
    int st;
    int ed;
    int th;
    MyLine() {}
    MyLine( int _st, int _ed, int _th ): st(_st), ed(_ed), th(_th) { }
};

vector<MyLine> hori;
vector<MyLine> vert;
bool ok[MAXN][MAXN];

int solved()
{
    int ans = 0;
    int lenH = hori.size();
    int lenV = vert.size();

    memset( ok, false, sizeof(ok) );

    for ( int i = 0; i < lenH; ++i )
    {
        for ( int j = 0; j < lenV; ++j )
            if ( vert[j].th >= hori[i].st && vert[j].th <= hori[i].ed
                && hori[i].th >= vert[j].st && hori[i].th <= vert[j].ed ) ok[i][j] = true;
    }


    for ( int i = 0; i < lenH; ++i )
    for ( int j = i + 1; j < lenH; ++j )
    {
        int cnt = 0;
        for ( int k = 0; k < lenV; ++k )
        if ( ok[i][k] && ok[j][k] ) ++cnt;
        ans += cnt * (cnt - 1) / 2;
    }

    return ans;
}

int main()
{
   // freopen( "input.txt", "r", stdin );
    int T;
    scanf( "%d", &T );
    while ( T-- )
    {
        int N;
        scanf( "%d", &N );
        int x1, y1, x2, y2;
        vert.clear();
        hori.clear();
        for ( int i = 0; i < N; ++i )
        {
            scanf( "%d%d%d%d", &x1, &y1, &x2, &y2 );
            if ( x1 == x2 )
            {
                if ( y1 > y2 ) swap( y1, y2 );
                hori.push_back( MyLine( y1, y2, x1 ) );
            }
            else if ( y1 == y2 )
            {
                if ( x1 > x2 ) swap( x1, x2 );
                vert.push_back( MyLine( x1, x2, y1 ) );
            }
        }

        printf( "%d\n", solved() );
    }
    return 0;
}