poj3168 Barn Expansion【计算几何 平面扫描】

Farmer John has N (1 <= N <= 25,000) rectangular barns on his farm, all with sides parallel to the X and Y axes and integer corner coordinates in the range 0..1,000,000. These barns do not overlap although they may share corners and/or sides with other barns. 

Since he has extra cows to milk this year, FJ would like to expand some of his barns. A barn has room to expand if it does not share a corner or a wall with any other barn. That is, FJ can expand a barn if all four of its walls can be pushed outward by at least some amount without bumping into another barn. If two barns meet at a corner, neither barn can expand. 

Please determine how many barns have room to expand.

Input

Line 1: A single integer, N 

Lines 2..N+1: Four space-separated integers A, B, C, and D, describing one barn. The lower-left corner of the barn is at (A,B) and the upper right corner is at (C,D).

Output

Line 1: A single integer that is the number of barns that can be expanded.

Sample Input

5
0 2 2 7
3 5 5 8
4 2 6 4
6 1 8 6
0 0 8 1

Sample Output

2

Hint

Explanation of the sample: 

There are 5 barns. The first barn has its lower-left corner at (0,2) and its upper-right corner at (2,7), and so on. 

Only two barns can be expanded --- the first two listed in the input. All other barns are each in contact with at least one other barn.

思路：

把四条边拆开 存到两个数组里

排序 y方向的先按照x排 再按照y方向上的起点排

x方向同理

遍历 对于每一个点 所有loc和他相同的 看看在不在重合范围内 并且更新范围

对于这个点本身的计数要特别一点 不然会重复

用cin cout会T 还是不长记性哦

可能真的痛经痛傻了 洗洗睡了洗洗睡了

#include <iostream>
#include <algorithm>
#include <cstring>

#include <cstdio>

#include <cmath>

using namespace std;
#define PI 3.1415926
#define EPS 1.0e-6



struct Point {
    Point(){}
    Point(double x, double y, double z):x(x), y(y), z(z){}
    double x,y, z;

};
struct rect{
    Point left_bottom;
    Point right_up;
};
struct barn{
    int index;
    int st, ed, loc;
}hh[500005], ll[500005];

bool cmp(barn a, barn b)
{
    if(a.loc == b.loc)
        return a.st < b.st;
    return a.loc < b.loc;
}




int n, ans;
rect rec[25005];
bool vis[25005];

void solve(barn *bar, int n)
{
    int i = 0;
    while(i < n){
        /*if(vis[bar[i].index]){
            i++;
            continue;
        }*/
        int pos = bar[i].loc;
        int cnt = 0;
        int first = i;
        int ed = bar[i].ed;
        i++;
        while(i < n && bar[i].loc == pos && bar[i].st <= ed){
            if(bar[i].ed > ed) ed = bar[i].ed;
            if(!vis[bar[i].index]){
                vis[bar[i].index] = true;
                ans--;
            }
            cnt++;
            i++;
        }
        if(cnt){
            if(!vis[bar[first].index]){
                vis[bar[first].index] = true;
                ans--;
            }

        }
    }
}

int main()
{
    //while(cin>>n){
    scanf("%d",&n);
        ans = n;
        //memset(vis, 0, sizeof(vis));
        for(int i = 0; i < n; i++){
            int left, bottom, right, up;
            scanf("%d%d%d%d",&left, &bottom, &right, &up);
            hh[2 * i].index = hh[2 * i + 1].index = i;
            ll[2 * i].index = ll[2 * i + 1].index = i;
            hh[2 * i].st = hh[2 * i + 1].st = bottom;
            hh[2 * i].ed = hh[2 * i + 1].ed = up;
            hh[2 * i].loc = left;hh[2 * i + 1].loc = right;
            ll[2 * i].st = ll[2 * i + 1].st = left;
            ll[2 * i].ed = ll[2 * i + 1].ed = right;
            ll[2 * i].loc = bottom; ll[2 * i + 1].loc = up;
        }

        sort(hh, hh + 2 * n, cmp);
        sort(ll, ll + 2 * n, cmp);
        solve(hh, 2 * n);
        solve(ll, 2 * n);
        //int ans = 0;
        //for(int i = 0; i < n; i++){
        //    if(!vis[i])ans++;
        //}
        printf("%d
", ans);
    //}
    return 0;

}