A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others. The length of the boundary of the union of all rectangles is called the perimeter.
Write a program to calculate the perimeter. An example with 7 rectangles is shown in Figure 1.
The corresponding boundary is the whole set of line segments drawn in Figure 2.
The vertices of all rectangles have integer coordinates.
Write a program to calculate the perimeter. An example with 7 rectangles is shown in Figure 1.
The corresponding boundary is the whole set of line segments drawn in Figure 2.
The vertices of all rectangles have integer coordinates.
Input
Your program is to read from standard input. The first line contains the number of rectangles pasted on the wall. In each of the subsequent lines, one can find the integer coordinates of the lower left vertex and the upper right vertex of each rectangle. The values of those coordinates are given as ordered pairs consisting of an x-coordinate followed by a y-coordinate.
0 <= number of rectangles < 5000
All coordinates are in the range [-10000,10000] and any existing rectangle has a positive area.
0 <= number of rectangles < 5000
All coordinates are in the range [-10000,10000] and any existing rectangle has a positive area.
Output
Your program is to write to standard output. The output must contain a single line with a non-negative integer which corresponds to the perimeter for the input rectangles.
Sample Input
7 -15 0 5 10 -5 8 20 25 15 -4 24 14 0 -6 16 4 2 15 10 22 30 10 36 20 34 0 40 16
Sample Output
228
这题写了很久很久
有点难受
现在还是有点迷糊
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <cmath> 5 #include <cstdio> 6 using namespace std; 7 typedef long long LL; 8 const int maxn = 5e4 + 10; 9 #define rtl rt<<1 10 #define rtr rt<<1|1 11 #define bug printf("*******"); 12 struct LINE { 13 int x, y1, y2, flag; 14 } line[maxn]; 15 int cmp(LINE a, LINE b) { 16 if (a.x == b.x) return a.flag > b.flag; 17 return a.x < b.x; 18 } 19 struct node { 20 int l, r, line, cover; 21 int m, lbd, rbd; 22 } tree[maxn << 2]; 23 int y[maxn]; 24 void build(int l, int r, int rt) { 25 tree[rt].l = l, tree[rt].r = r; 26 tree[rt].cover = tree[rt].m = tree[rt].line = 0; 27 tree[rt].lbd = tree[rt].rbd = 0; 28 if (r - l > 1) { 29 int m = (l + r) >> 1; 30 build(l, m, rtl); 31 build(m, r, rtr); 32 } 33 } 34 void update_line(int rt) { 35 if (tree[rt].cover > 0) { 36 tree[rt].lbd = tree[rt].rbd = tree[rt].line = 1; 37 } else if (tree[rt].r - tree[rt].l == 1) { 38 tree[rt].lbd = tree[rt].rbd = tree[rt].line = 0; 39 } else { 40 tree[rt].lbd = tree[rtl].lbd; 41 tree[rt].rbd = tree[rtr].rbd; 42 tree[rt].line = tree[rtl].line + tree[rtr].line - tree[rtl].rbd * tree[rtr].lbd; 43 } 44 } 45 void update_m(int rt) { 46 if (tree[rt].cover > 0) tree[rt].m = y[tree[rt].r] - y[tree[rt].l]; 47 else if (tree[rt].r - tree[rt].l == 1) tree[rt].m = 0; 48 else tree[rt].m = tree[rtl].m + tree[rtr].m; 49 50 } 51 void add(int l, int r, int rt) { 52 if (y[tree[rt].l] >= l && y[tree[rt].r] <= r) tree[rt].cover++; 53 else if (tree[rt].r - tree[rt].l == 1) return ; 54 else { 55 int m = (tree[rt].l + tree[rt].r) >> 1; 56 if (r <= y[m]) add(l, r, rtl); 57 else if (l > y[m]) add(l, r, rtr); 58 else { 59 add(l, y[m], rtl); 60 add(y[m], r, rtr); 61 } 62 } 63 update_line(rt); 64 update_m(rt); 65 } 66 void del(int l, int r, int rt) { 67 if (y[tree[rt].l] >= l && y[tree[rt].r] <= r) tree[rt].cover--; 68 else if (tree[rt].r - tree[rt].l == 1) return; 69 else { 70 int m = (tree[rt].l + tree[rt].r) >> 1; 71 if (r <= y[m]) del(l, r, rtl); 72 else if (l > y[m]) del(l, r, rtr); 73 else { 74 del(l, y[m], rtl); 75 del(y[m], r, rtr); 76 } 77 } 78 update_line(rt); 79 update_m(rt); 80 } 81 int main() { 82 int n, cnt = 0; 83 scanf("%d", &n); 84 for (int i = 0 ; i < n ; i++) { 85 int x1, y1, x2, y2; 86 scanf("%d%d%d%d", &x1, &y1, &x2, &y2); 87 line[cnt].x = x1; 88 line[cnt].y1 = y1; 89 line[cnt].y2 = y2; 90 line[cnt].flag = 1; 91 y[cnt++] = y1; 92 line[cnt].x = x2; 93 line[cnt].y1 = y1; 94 line[cnt].y2 = y2; 95 line[cnt].flag = 0; 96 y[cnt++] = y2; 97 } 98 sort(y, y + cnt); 99 sort(line, line + cnt, cmp); 100 int len = unique(y, y + cnt) - y; 101 build(0, len - 1, 1); 102 int ans = 0, now_m = 0, now_line = 0; 103 for (int i = 0 ; i < cnt ; i++) { 104 if (line[i].flag) add(line[i].y1, line[i].y2, 1); 105 else del(line[i].y1, line[i].y2, 1); 106 if (i >= 1) ans += 2 * now_line * (line[i].x - line[i - 1].x); 107 ans += abs(tree[1].m - now_m); 108 now_m = tree[1].m; 109 now_line = tree[1].line; 110 } 111 printf("%d ", ans); 112 return 0; 113 }