Atlantis
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 16222 | Accepted: 6172 |
Description
There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.
Input
The input consists of several test cases. Each test case starts with a line containing a single integer n (1 <= n <= 100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0 <= x1 < x2 <= 100000;0 <= y1 < y2 <= 100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.
The input file is terminated by a line containing a single 0. Don't process it.
The input file is terminated by a line containing a single 0. Don't process it.
Output
For each test case, your program should output one section. The first line of each section must be "Test case #k", where k is the number of the test case (starting with 1). The second one must be "Total explored area: a", where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.
Output a blank line after each test case.
Output a blank line after each test case.
Sample Input
2 10 10 20 20 15 15 25 25.5 0
Sample Output
Test case #1 Total explored area: 180.00
思路:首先将横坐标离散化,再对纵坐标排序,然后根据y轴从下往上扫描,每次的高度就是seg[i].y-seg[i-1].y,这就相当于分矩形的宽,然后要做的事就是查询x轴(矩形长)的有效长度,这就要交给线段树了。
AC代码:
1 /************************************************************************* 2 > File Name: area.cpp 3 > Author: wangzhili 4 > Mail: wangstdio.h@gmail.com 5 > Created Time: 2014/3/1 星期六 16:06:57 6 ************************************************************************/ 7 8 #include<iostream> 9 #include<algorithm> 10 #include<cstdio> 11 #define MAX 1000 12 using namespace std; 13 class TreeNode 14 { 15 public: 16 int left; 17 int right; 18 int mid; 19 int cover; 20 int flag; 21 double length; 22 }; 23 typedef struct 24 { 25 double xl, xr, y; 26 int flag; 27 }Line; 28 TreeNode node[3*MAX]; 29 Line seg[MAX]; 30 double x[MAX]; 31 double length; 32 bool cmp(Line a, Line b) 33 { 34 return a.y < b.y; 35 } 36 37 void BuildTree(int k, int l, int r) 38 { 39 node[k].left = l; 40 node[k].right = r; 41 node[k].mid = (l + r) >> 1; 42 node[k].cover = 0; 43 node[k].flag = 0; 44 if(l + 1 == r) 45 { 46 node[k].flag = 1; 47 return ; 48 } 49 int mid = (l + r) >> 1; 50 BuildTree(k << 1, l, mid); 51 BuildTree(k << 1|1, mid, r); 52 } 53 54 void UpdateTree(int k, int l, int r, int flag) 55 { 56 if(node[k].left == l && node[k].right == r) 57 { 58 node[k].cover += flag; 59 node[k].length = x[r-1] - x[l-1]; 60 return ; 61 } 62 if(node[k].flag) 63 return ; 64 if(node[k].mid <= l) 65 UpdateTree(k << 1|1, l, r, flag); 66 else if(node[k].mid >= r) 67 UpdateTree(k << 1, l, r, flag); 68 else 69 { 70 UpdateTree(k << 1, l, node[k].mid, flag); 71 UpdateTree(k << 1|1, node[k].mid, r, flag); 72 } 73 } 74 75 void GetLength(int k) 76 { 77 if(node[k].cover > 0) 78 { 79 length += node[k].length; 80 return ; 81 } 82 if(node[k].flag) 83 return; 84 GetLength(k << 1); 85 GetLength(k << 1|1); 86 } 87 88 int GetIndex(double num, int length) 89 { 90 int l, r, mid; 91 l = 0, r = length-1; 92 while(l <= r) 93 { 94 mid = (l + r) >> 1; 95 if(x[mid] == num) 96 return mid; 97 else if(x[mid] > num) 98 r = mid - 1; 99 else 100 l = mid + 1; 101 } 102 } 103 104 int main(int argc, char const *argv[]) 105 { 106 int n, i, j, k, cnt; 107 int xl, xr; 108 double ans; 109 double x1, y1, x2, y2; 110 cnt = 0; 111 BuildTree(1, 1, 205); 112 // freopen("in.c", "r", stdin); 113 while(~scanf("%d", &n) && n) 114 { 115 j = 0; 116 ans = 0.; 117 for(i = 1; i < 408; i ++) 118 { 119 node[i].cover = 0; 120 } 121 for(i = 0; i < n; i ++) 122 { 123 scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2); 124 seg[j].xl = x1; 125 seg[j].xr = x2; 126 seg[j].y = y1; 127 x[j] = x1; 128 seg[j ++].flag = 1; 129 seg[j].xl = x1; 130 seg[j].xr = x2; 131 seg[j].y = y2; 132 x[j] = x2; 133 seg[j ++].flag = -1; 134 } 135 sort(x, x+j); 136 sort(seg, seg+j, cmp); 137 k = 1; 138 for(i = 1; i < j; i ++) 139 { 140 if(x[i] != x[i-1]) 141 x[k ++] = x[i]; 142 } 143 xl = GetIndex(seg[0].xl, k) + 1; 144 xr = GetIndex(seg[0].xr, k) + 1; 145 UpdateTree(1, xl, xr, seg[0].flag); 146 length = 0; 147 GetLength(1); 148 for(i = 1; i < j; i ++) 149 { 150 ans += (seg[i].y-seg[i-1].y)*length; 151 xl = GetIndex(seg[i].xl, k)+1; 152 xr = GetIndex(seg[i].xr, k)+1; 153 UpdateTree(1, xl, xr, seg[i].flag); 154 length = 0.; 155 GetLength(1); 156 } 157 printf("Test case #%d Total explored area: %.2lf ", ++cnt, ans); 158 } 159 return 0; 160 }