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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): 4823 Accepted Submission(s): 2398 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
/*hdu 1255 覆盖的面积(求覆盖至少两次以上的面积)主要是在push_up那出现了问题设定len是覆盖两次以上的长度，len1是覆盖1次以上的长度那么有以下几种情况：1.sum>1 则说当前区间被覆盖至少两次2.l == r 直接为03.当前区间一次都没被覆盖过 lson.len + rson.len4.当前区间被覆盖过一次sum=1,那么 lson.len1 + rson.len1，即左右孩子中被 //最开始没有考虑到这个情况TAT至少覆盖过一次的和hhh-2016-03-27 18:07:36*/#include#include #include #include #include using namespace std;#define lson (i<<1)#define rson ((i<<1)|1)typedef long long ll;const int maxn = 50050;struct node{ int l,r,sum; double len; double len1; int mid() { return (l+r)>>1; }} tree[maxn*5];double hs[maxn];int tot,m;void push_up(int i){ if(tree[i].sum) tree[i].len1 = hs[tree[i].r+1]-hs[tree[i].l]; else if(tree[i].l == tree[i].r) tree[i].len1 = 0; else tree[i].len1 = tree[lson].len1+tree[rson].len1; if(tree[i].sum > 1) tree[i].len = hs[tree[i].r+1]-hs[tree[i].l]; else if(tree[i].l == tree[i].r) tree[i].len = 0; else if(tree[i].sum == 1) tree[i].len = tree[lson].len1 + tree[rson].len1; else tree[i].len = tree[lson].len+tree[rson].len;}void build(int i,int l,int r){ tree[i].l = l,tree[i].r = r; tree[i].sum=tree[i].len= 0; tree[i].len1 = 0; if(l ==r ) { return ; } int mid=tree[i].mid(); build(lson,l,mid); build(rson,mid+1,r); push_up(i);}void push_down(int i){}void Insert(int i,int l,int r,int val){ if(tree[i].l >= l && tree[i].r <= r) { tree[i].sum += val; push_up(i); return; } push_down(i); int mid = tree[i].mid(); if(l <= mid) Insert(lson,l,r,val); if(r > mid) Insert(rson,l,r,val); push_up(i);}struct edge{ double l,r,high; int va; edge() {}; edge(double ll,double rr,double h,int v):l(ll),r(rr),high(h),va(v) {}} tx[maxn];bool cmp(edge a,edge b){ if(a.high != b.high) return a.high < b.high; else return a.va > b.va;}int fin(double x){ int l = 0,r = m-1; while(l <= r) { int mid = (l+r)>>1; if(hs[mid] == x) return mid; else if(hs[mid] < x) l = mid+1; else r = mid-1; }}int main(){ int n,x,q,t; int cas =1; scanf("%d",&t); while(t--) { scanf("%d",&n); tot = 0; double x1,x2,y1,y2; for(int i = 1; i <= n; i++) { scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); hs[tot] = x1; tx[tot++] = edge(x1,x2,y1,1); hs[tot] = x2; tx[tot++] = edge(x1,x2,y2,-1); } sort(hs,hs+tot); sort(tx,tx+tot,cmp); m = 1; for(int i = 1; i < tot; i++) if(hs[i] != hs[i-1]) hs[m++] = hs[i]; build(1,0,m-1); double ans = 0; for(int i = 0;i < tot;i++) { int l = fin(tx[i].l); int r = fin(tx[i].r)-1; //cout << tx[i].l <<" " << tx[i].r <<" "<