【HDU1255】【线段树】【扫描线】【面积】

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覆盖的面积

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4417 Accepted Submission(s): 2180

Problem Description

给定平面上若干矩形,求出被这些矩形覆盖过至少两次的区域的面积.

Input

输入数据的第一行是一个正整数T(1<=T<=100),代表测试数据的数量.每个测试数据的第一行是一个正整数N(1<=N<=1000),代表矩形的数量,然后是N行数据,每一行包含四个浮点数,代表平面上的一个矩形的左上角坐标和右下角坐标,矩形的上下边和X轴平行,左右边和Y轴平行.坐标的范围从0到100000.

注意:本题的输入数据较多,推荐使用scanf读入数据.

Output

对于每组测试数据,请计算出被这些矩形覆盖过至少两次的区域的面积.结果保留两位小数.

Sample Input

251 1 4 21 3 3 72 1.5 5 4.53.5 1.25 7.5 46 3 10 730 0 1 11 0 2 12 0 3 1

Sample Output

7.630.00

Author

Ignatius.L & weigang Lee

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Ignatius.L

#include <iostream>#include <cstring>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <string>#include <cstdlib>#include <cstdio>#include <algorithm>using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define mp push_back#define lson l,m,rt<<1#define rson m,r,rt<<1|1int T;int n;struct Edge{double l,r,y;int f;Edge(){};Edge(double ll,double rr,double yy,int ff):l(ll),r(rr),y(yy),f(ff){};}edge[1010*4];int len;double X[1010*4];double sum[1010*10];double more[1010*10];int col[1010*10];int cmp(Edge a,Edge b){return a.y < b.y;}void PushUp(int rt,int l,int r){if(col[rt] >= 2){sum[rt] = more[rt] = X[r]- X[l];}else if(col[rt] == 1){sum[rt] = X[r] - X[l];if(r - l!=1)more[rt] = sum[rt<<1] + sum[rt<<1|1];elsemore[rt] = 0;}else{if(r - l == 1){sum[rt] = more[rt] = 0;}else{sum[rt] = sum[rt<<1] + sum[rt<<1|1];more[rt] = more[rt<<1] + more[rt<<1|1];}}}void update(double L,double R,int c,int l,int r,int rt){if(L <= X[l] && X[r] <= R ){col[rt] += c;PushUp(rt,l,r);return;}int m = (l + r) >> 1;if(L < X[m]) update(L,min(R,X[m]),c,lson);if(X[m] < R) update(max(L,X[m]),R,c,rson);PushUp(rt,l,r);}int main(){scanf("%d",&T);while(T --){len = 0;scanf("%d",&n);for(int i=0;i<n;i++){double x1,y1,x2,y2;scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);if(x1 > x2) swap(x1,x2);if(y1 > y2) swap(y1,y2);edge[len] = Edge(x1,x2,y1,1);X[len++] = x1;edge[len] = Edge(x1,x2,y2,-1);X[len++] = x2;}sort(edge,edge+len,cmp);sort(X,X+len);int MAXN = 1;for(int i=1;i<len;i++){if(X[i] != X[i-1]){X[MAXN++] = X[i]; }}//build(0,MAXN-1,1);memset(more,0,sizeof(more));memset(sum,0,sizeof(sum));memset(col,0,sizeof(col));double ans = 0;for(int i=0;i<len-1;i++){update(edge[i].l,edge[i].r,edge[i].f,0,MAXN-1,1);ans += more[1] * (edge[i+1].y - edge[i].y);}printf("%.2lf\n",ans);}}

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