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
这题也是模板提
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <cmath> 5 #include <ctype.h> 6 #include <set> 7 #include <map> 8 #include <queue> 9 #include <stack> 10 #include <iostream> 11 using namespace std; 12 #define bug printf("****** "); 13 #define rtl rt<<1 14 #define rtr rt<<1|1 15 typedef long long LL; 16 const int maxn = 1e5 + 10; 17 struct LINE { 18 double x, y1, y2; 19 int flag; 20 } line[205]; 21 int cmp(LINE a, LINE b) { 22 return a.x < b.x; 23 } 24 struct node { 25 double pre, l, r; 26 int cover, flag; 27 } tree[1210]; 28 double y[205]; 29 void build(int rt, int l, int r) { 30 tree[rt].l = y[l], tree[rt].r = y[r]; 31 tree[rt].flag = -1, tree[rt].cover = 0, tree[rt].pre = -1; 32 if (l + 1 == r) { 33 tree[rt].flag = 1; 34 return ; 35 } 36 int m = (l + r) >> 1; 37 build(rtl, l, m); 38 build(rtr, m, r); 39 } 40 double query(int rt, double x, double y1, double y2, int flag) { 41 if (tree[rt].r <= y1 || tree[rt].l >= y2) return 0; 42 if (tree[rt].flag == 1) { 43 if (tree[rt].cover > 0) { 44 double pre = tree[rt].pre; 45 double ans = (x - pre) * (tree[rt].r - tree[rt].l); 46 tree[rt].pre = x; 47 tree[rt].cover += flag; 48 return ans; 49 } else { 50 tree[rt].cover += flag; 51 tree[rt].pre = x; 52 return 0; 53 } 54 } 55 return query(rtl, x, y1, y2, flag) + query(rtr, x, y1, y2, flag); 56 } 57 int main() { 58 int cas = 1, n; 59 while(scanf("%d", &n), n) { 60 int cnt = -1; 61 for (int i = 0 ; i < n ; i++) { 62 double x1, y1, x2, y2; 63 scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2); 64 y[++cnt] = y1; 65 line[cnt].flag = 1; 66 line[cnt].x = x1; 67 line[cnt].y1 = y1; 68 line[cnt].y2 = y2; 69 y[++cnt] = y2; 70 line[cnt].flag = -1; 71 line[cnt].x = x2; 72 line[cnt].y1 = y1; 73 line[cnt].y2 = y2; 74 } 75 sort(y, y + cnt + 1); 76 sort(line, line + cnt + 1, cmp); 77 build(1, 0, cnt); 78 double ans = 0; 79 for (int i = 0 ; i <= cnt ; i++ ) 80 ans += query(1, line[i].x, line[i].y1, line[i].y2, line[i].flag); 81 printf("Test case #%d ", cas++); 82 printf("Total explored area: %.2lf ", ans); 83 } 84 return 0; 85 }