POJ -- 1151

时间：2014-03-02 14:54:07 阅读：295 评论：0 收藏：0 [点我收藏+]

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.

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.

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\nTotal explored area: %.2lf\n\n", ++cnt, ans); 
158     }
159     return 0; 
160 }

POJ -- 1151,布布扣,bubuko.com

POJ -- 1151

原文：http://www.cnblogs.com/anhuizhiye/p/3575398.html

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