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POJ 1151 Atlantis 矩形切割 || 线段树 扫描线

热度:34   发布时间:2024-01-13 18:13:02.0

这道题的题意就是给出一些矩形,问这些矩形覆盖的面积,也就是矩形可能是相交的

这道题由于数据量很小,所以可以按照将输入的坐标排序的方法进行分割矩形

如下面的代码

/*
ID: sdj22251
PROG: subset
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <map>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAXN 15500
#define INF 1000000000
using namespace std;
double a1[122], b1[122], a2[122], b2[122];
double t1[222], t2[222];
int main()
{int n, cnt, cas = 0;while(scanf("%d", &n) != EOF && n){cnt = 0;for(int i = 0; i < n; i++){scanf("%lf%lf%lf%lf", &a1[i], &b1[i], &a2[i], &b2[i]);t1[cnt] = a1[i];t2[cnt] = b1[i];cnt++;t1[cnt] = a2[i];t2[cnt] = b2[i];cnt++;}sort(t1, t1 + cnt);sort(t2, t2 + cnt);double ans = 0;for(int i = 0; i < cnt - 1; i++){for(int j = 0; j < cnt - 1; j++){for(int k = 0; k < n; k++){if(t1[i] >= a1[k] && t2[j] >= b1[k] && t1[i + 1] <= a2[k] && t2[j + 1] <= b2[k]){ans += (t1[i + 1] - t1[i]) * (t2[j + 1] - t2[j]);break;}}}}printf("Test case #%d\n", ++cas);printf("Total explored area: %.2f\n\n", ans);}return 0;
}

还有一种就是矩形切割了,其原理就是将相交的矩形拆分成几个小的矩形,然后最后所有的矩形都不会相交,求出面积和即可

矩形切割    根据线段切割的思想,我们稍做推广,便能得出矩形切割的方法。

类似地,若矩形集合中已有矩形(x1,y1,x2,y2),现加入矩形(x3,y3,x4,y4)。它们的位置关系可以有很多种(有17种之多),这里就不一一列举了。但无论它们的位置关系如何复杂,运用线段切割的思想来进行矩形切割,就会变得十分明了。

我们将矩形的切割正交分解,先进行x方向上的切割,再进行y方向的切割。

以下图为例:

 

插入矩形(x3,y3,x4,y4)后,对矩形(x1,y1,x2,y2)进行切割。

Step 1:首先从x方向上切。把线段(x1,x2)切成(x1,x3),(x4,x2)两条线段。于是相应地,我们就把两个矩形切了出来——(x1,y1,x3,y2),(x4,y1,x2,y2)。把它们加到矩形集合中。去掉了这两个矩形后,我们要切的矩形就变为(x3,y1,x4,y2)。

Step 2:接着我们再进行y方向上的切割。把线段(y1,y2)切成(y1,y3)。相应地又得到一个矩形(x3,y1,x4,y2)。把它放入矩形集合。

Step 3:剩下的矩形为(x3,y3,x4,y2),这个矩形已经被矩形(x3,y3,x4,y4)覆盖了,因此直接把它删掉。

 

我们可以归纳出矩形切割的思想

1、先对被切割矩形进行x方向上的切割。取(x1,x2),(x3,x4)的交集(k1,k2)

① 若x1<k1,则加入矩形(x1,y1,k1,y2)

② 若k2<x2,则加入矩形(k2,y1,x2,y2)

2、再对切剩的矩形(k1,y1,k2,y2) 进行y方向上的切割。取(y1,y2),(y3,y4)的交集(k3,k4)

① 若y1<k3,则加入矩形(k1,y1,k2,k3)

② 若k4<y2,则加入矩形(k1,k4,k2,y2)

3、把矩形(x1,y1,x2,y2)从矩形集合中删除。


然后实现代码如下

/*
ID: sdj22251
PROG: rect1
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAXN 2555
#define MAXM 104444
#define INF 100000000
#define eps 1e-7
#define L(X) X<<1
#define R(X) X<<1|1
using namespace std;
int n;
double xa[1010], xb[1010], ya[1010], yb[1010], area;
void cover(double lx, double rx, double ly, double ry, int t)
{while((t <= n) && ((lx >= xb[t]) || (rx <= xa[t]) || (ly >= yb[t]) || (ry <= ya[t])))t++;if(t > n){area += (rx - lx ) * (ry - ly);return;}if(lx < xa[t]){cover(lx, xa[t], ly, ry, t + 1 );lx = xa[t];}if(rx > xb[t]){cover(xb[t], rx , ly, ry, t + 1 );rx = xb[t];}if(ly < ya[t]){cover(lx, rx, ly, ya[t], t + 1 );ly = ya[t];}if(ry > yb[t]){cover(lx, rx, yb[t], ry, t + 1 );ry = yb[t];}
}
int main()
{int cas = 0;while(scanf("%d", &n) != EOF && n){area = 0;for(int i = 1; i <= n; i++)scanf("%lf%lf%lf%lf", &xa[i], &ya[i], &xb[i], &yb[i]);for(int i = n; i >= 1; i--)cover(xa[i], xb[i], ya[i], yb[i], i + 1);printf("Test case #%d\n", ++cas);printf("Total explored area: %.2f\n\n", area);}return 0;
}


然后放出线段树版本的


/*
ID: sdj22251
PROG: subset
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAXN 2222
#define MAXM 164444
#define INF 100000000
#define eps 1e-7
#define L(X) X<<1
#define R(X) X<<1|1
using namespace std;struct node
{int left, right, mid;int cnt;double sum;
}tree[4 * MAXN];struct Seg
{double h, l, r;int s;Seg(){}Seg(double a, double b, double c, int d) {l = a; r = b; h = c; s = d;}bool operator <(const Seg &cmp)const{return h < cmp.h;}
}seg[MAXN];
double x[MAXN];void up(int C)
{if(tree[C].cnt) tree[C].sum = x[tree[C].right + 1] - x[tree[C].left];else if(tree[C].left == tree[C].right) tree[C].sum = 0;else tree[C].sum = tree[L(C)].sum + tree[R(C)].sum;
}void make_tree(int s, int e, int C)
{tree[C].left = s;tree[C].right = e;tree[C].mid = (s + e) >> 1;tree[C].cnt = 0;tree[C].sum = 0;if(s == e) return;make_tree(s, tree[C].mid, L(C));make_tree(tree[C].mid + 1, e, R(C));
}void update(int s, int e, int v, int C)
{if(tree[C].left >= s && tree[C].right <= e){tree[C].cnt += v;up(C);return;}if(tree[C].mid >= s) update(s, e, v, L(C));if(tree[C].mid < e) update(s, e, v, R(C));up(C);
}int bin(double v, int bound)
{int low = 0, high = bound - 1;while(low <= high){int mid = (low + high) >> 1;if(x[mid] == v) return mid;if(x[mid] < v) low = mid + 1;else high = mid - 1;}return -1;
}int main()
{int n, cas = 0;while(scanf("%d", &n) != EOF && n){int m = 0;double a, b, c, d;while(n--){scanf("%lf%lf%lf%lf", &a, &b, &c, &d);x[m] = a;seg[m++] = Seg(a, c, b, 1);x[m] = c;seg[m++] = Seg(a, c, d, -1);}sort(x, x + m);sort(seg, seg + m);int k = unique(x, x + m) - x;make_tree(0, k - 1, 1);double ans = 0;for(int i = 0; i < m - 1; i ++){int l = bin(seg[i].l, k);int r = bin(seg[i].r, k) - 1;  //这里-1的原因是建树时是建到叶子结点的,而我们更新的是线段update(l, r, seg[i].s, 1);ans += tree[1].sum * (seg[i + 1].h - seg[i].h);}printf("Test case #%d\nTotal explored area: %.2f\n\n", ++cas, ans);}return 0;
}