[ZOJ1426 ][POJ1693] Counting Rectangles

ZOJ: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=426

POJ: http://poj.org/problem?id=1693

题目大意：

给出一些水平和竖直线段，问围城多少矩形。

注意简化条件：

 The input line segments are such that each intersection point comes from the intersection of exactly one horizontal segment and one vertical segment.

解题思路：

据说可以暴力，就暴力了……很久没有1AC了……ZOJ交完交POJ交错题了……

源代码：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 110 

typedef struct line_ver
{
    int x, y1, y2;
}line_var;

typedef struct line_hor
{
    int y, x1, x2;
}line_hor;

int v,h,n;
line_ver linev[maxn];
line_hor lineh[maxn];

int min(int x, int y){if (x>y) return y; return x;}
int max(int x, int y){if (x>y) return x; return y;}

void init()
{
    int x1,y1,x2,y2;
    scanf("%d", &n);
    v=h=0;
    for (int i=0; i<n; i++)
    {
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        if (x1==x2)     //vertical
        {
            linev[v].x=x1; linev[v].y1=min(y1,y2); linev[v++].y2=max(y1,y2);
        }
        else            //horizontal
        {
            lineh[h].y=y1; lineh[h].x1=min(x1,x2); lineh[h++].x2=max(x1,x2);
        }
    }
}

bool intersect(int vv, int hh)
{
    if (linev[vv].x>=lineh[hh].x1 && linev[vv].x<=lineh[hh].x2)
        if (lineh[hh].y>=linev[vv].y1 && lineh[hh].y<=linev[vv].y2)
            return true;
    return false;
}

int solve()
{
    int ret=0;
    for (int v1=0; v1<v-1; v1++)
        for (int h1=0; h1<h-1; h1++)
            if (intersect(v1, h1))
                for (int v2=v1+1; v2<v; v2++)
                    if (intersect(v2, h1))
                        for (int h2=h1+1; h2<h; h2++)
                            if (intersect(v1, h2) && intersect(v2, h2))
                                ret++;
    return ret;                            
}

int main()
{
    int cs, ans;
    scanf("%d", &cs);
    while (cs--)
    {
        init();
        ans=solve();
        printf("%d\n", ans);
    }
}