HDU1542 Atlantis —— 求矩形面积并 线段树 + 扫描线 + 离散化

Posted 2020-10-12 h_z_cong

题目链接：https://vjudge.net/problem/HDU-1542

 

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. 

InputThe input file 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.OutputFor 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 

 

 

题解：

 

 

写法一：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 using namespace std;
13 typedef long long LL;
14 const double EPS = 1e-8;
15 const int INF = 2e9;
16 const LL LNF = 2e18;
17 const int MAXN = 1e5+10;
18 
19 struct line
20 {
21     double le, ri, h;
22     int id;
23     bool operator<(const line &a){
24         return h<a.h;
25     }
26 }Line[MAXN];
27 
28 double sum[MAXN<<2], addv[MAXN<<2];
29 double X[MAXN];
30 
31 void push_up(int u, int l, int r)
32 {
33     if(addv[u]>=1) sum[u] = X[r] - X[l];
34     else sum[u] = (l+1==r)?0:sum[u*2]+sum[u*2+1];
35 }
36 
37 void add(int u, int l, int r, int x, int y, int v)
38 {
39     if(x<=l && r<=y)
40     {
41         addv[u] += v;
42         push_up(u, l, r);
43         return;
44     }
45 
46     int mid = (l+r)>>1;
47     if(x<=mid-1) add(u*2, l, mid, x, y, v);
48     if(y>=mid+1) add(u*2+1, mid, r, x, y, v);
49     push_up(u, l, r);
50 }
51 
52 int main()
53 {
54     int n, kase = 0;
55     while(scanf("%d", &n) && n)
56     {
57         for(int i = 1; i<=n; i++)
58         {
59             double x1, y1, x2, y2;
60             scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
61             Line[i].le = Line[i+n].le = x1;
62             Line[i].ri = Line[i+n].ri = x2;
63             Line[i].h = y1; Line[i+n].h = y2;
64             Line[i].id = 1; Line[i+n].id = -1;
65             X[i] = x1; X[i+n] = x2;
66         }
67 
68         sort(Line+1, Line+1+2*n);
69         sort(X+1, X+1+2*n);
70         int m = unique(X+1, X+1+2*n) - (X+1);
71 
72         memset(sum, 0, sizeof(sum));
73         memset(addv, 0, sizeof(addv));
74         double ans = 0;
75         for(int i = 1; i<=2*n-1; i++)
76         {
77             int l = upper_bound(X+1, X+1+m, Line[i].le) - (X+1);
78             int r = upper_bound(X+1, X+1+m, Line[i].ri) - (X+1);
79             add(1, 1, m, l, r, Line[i].id);
80             ans += sum[1]* (Line[i+1].h-Line[i].h);
81         }
82         printf("Test case #%d\n", ++kase);
83         printf("Total explored area: %.2f\n\n", ans);
84     }
85 }

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写法二：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 using namespace std;
13 typedef long long LL;
14 const double EPS = 1e-8;
15 const int INF = 2e9;
16 const LL LNF = 2e18;
17 const int MAXN = 1e5+10;
18 
19 struct line
20 {
21     double le, ri, h;
22     int id;
23     bool operator<(const line &a){
24         return h<a.h;
25     }
26 }Line[MAXN];
27 
28 double sum[MAXN<<2], addv[MAXN<<2];
29 double X[MAXN];
30 
31 void push_up(int u, int l, int r)   //区间左闭右开
32 {
33     if(addv[u]) sum[u] = X[r+1] - X[l];
34     else  sum[u] = (l==r)?0:(sum[u*2]+sum[u*2+1]);
35 }
36 
37 void add(int u, int l, int r, int x, int y, int v)
38 {
39     if(x<=l && r<=y)
40     {
41         addv[u] += v;
42         push_up(u, l, r);
43         return;
44     }
45 
46     int mid = (l+r)>>1;
47     if(x<=mid) add(u*2, l, mid, x, y, v);
48     if(y>=mid+1) add(u*2+1, mid+1, r, x, y, v);
49     push_up(u, l, r);
50 }
51 
52 int main()
53 {
54     int n, kase = 0;
55     while(scanf("%d", &n) && n)
56     {
57         for(int i = 1; i<=n; i++)
58         {
59             double x1, y1, x2, y2;
60             scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
61             Line[i].le = Line[i+n].le = x1;
62             Line[i].ri = Line[i+n].ri = x2;
63             Line[i].h = y1; Line[i+n].h = y2;
64             Line[i].id = 1; Line[i+n].id = -1;
65             X[i] = x1; X[i+n] = x2;
66         }
67 
68         sort(Line+1, Line+1+2*n);
69         sort(X+1, X+1+2*n);
70         int m = unique(X+1, X+1+2*n) - (X+1);
71 
72         memset(sum, 0, sizeof(sum));
73         memset(addv, 0, sizeof(addv));
74         double ans = 0;
75         for(int i = 1; i<=2*n-1; i++)
76         {
77             int l = upper_bound(X+1, X+1+m, Line[i].le) - (X+1);
78             int r = upper_bound(X+1, X+1+m, Line[i].ri) - (X+1);
79             add(1, 1, m, l, r-1, Line[i].id);
80             ans += sum[1]*(Line[i+1].h-Line[i].h);
81         }
82         printf("Test case #%d\n", ++kase);
83         printf("Total explored area: %.2f\n\n", ans);
84     }
85 }

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