覆盖的面积 HDU - 1255
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
2 5 1 1 4 2 1 3 3 7 2 1.5 5 4.5 3.5 1.25 7.5 4 6 3 10 7 3 0 0 1 1 1 0 2 1 2 0 3 1
Sample Output
7.63 0.00
和求面积并差不多。
1 #pragma warning(disable:4996) 2 #include<cstdio> 3 #include<stack> 4 #include<cmath> 5 #include<string> 6 #include<cstring> 7 #include<iostream> 8 #include<algorithm> 9 using namespace std; 10 11 #define lson root<<1 12 #define rson root<<1|1 13 #define ll long long 14 15 const int maxn = 5000; 16 17 struct Seg { 18 int d; 19 double l, r, h; 20 Seg(){} 21 Seg(double l, double r, double h, int d) : l(l), r(r), h(h), d(d) {} 22 bool operator<(const Seg& rhs) { return h < rhs.h; } 23 }line[maxn]; 24 25 int n, cnt[maxn]; 26 double Hash[maxn], one[maxn], two[maxn]; 27 28 void Push_up(int l, int r, int root) { 29 if (cnt[root] >= 2) two[root] = one[root] = Hash[r] - Hash[l]; 30 else if (cnt[root] == 1) { 31 one[root] = Hash[r] - Hash[l]; 32 if (l + 1 == r) two[root] = 0; 33 else two[root] = one[lson] + one[rson]; 34 } 35 else { 36 if (l + 1 == r) one[root] = two[root] = 0; 37 else { 38 one[root] = one[lson] + one[rson]; 39 two[root] = two[lson] + two[rson]; 40 } 41 } 42 } 43 44 void Update(int L, int R, int l, int r, int root, int c) { 45 if (l >= R || r <= L) return; 46 if (L <= l && r <= R) { 47 cnt[root] += c; 48 Push_up(l, r, root); 49 return; 50 } 51 int mid = (l + r) >> 1; 52 Update(L, R, l, mid, lson, c); 53 Update(L, R, mid, r, rson, c); 54 Push_up(l, r, root); 55 } 56 57 int main() 58 { 59 int kase; 60 scanf("%d", &kase); 61 while (kase--) { 62 scanf("%d", &n); 63 double x1, y1, x2, y2; 64 for (int i = 1; i <= n; i++) { 65 scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2); 66 line[i] = Seg(x1, x2, y1, 1); 67 Hash[i] = x1; 68 line[i + n] = Seg(x1, x2, y2, -1); 69 Hash[i + n] = x2; 70 } 71 72 n <<= 1; 73 sort(line + 1, line + n + 1); 74 sort(Hash + 1, Hash + n + 1); 75 76 int m = unique(Hash + 1, Hash + n + 1) - (Hash + 1); 77 78 memset(cnt, 0, sizeof(cnt)); 79 memset(one, 0, sizeof(one)); 80 memset(two, 0, sizeof(two)); 81 82 double ans = 0; 83 for (int i = 1; i <= n; i++) { 84 int l = lower_bound(Hash + 1, Hash + m + 1, line[i].l) - Hash; 85 int r = lower_bound(Hash + 1, Hash + m + 1, line[i].r) - Hash; 86 Update(l, r, 1, m, 1, line[i].d); 87 ans += two[1] * (line[i + 1].h - line[i].h); 88 } 89 printf("%.2lf\n", ans); 90 } 91 }
