描述
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.输入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.输出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.样例输入
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
样例输出
228
来源
IOI 1998
1 #include <iostream> 2 #include <string.h> 3 #include <algorithm> 4 #include <stack> 5 #include <string> 6 #include <math.h> 7 #include <queue> 8 #include <stdio.h> 9 #include <string.h> 10 #include <vector> 11 #include <fstream> 12 #include <set> 13 #include<map> 14 #define l(rt) rt*2 15 #define r(rt) rt*2+1 16 #define mid(l,r) (l+r)/2 17 18 using namespace std; 19 20 const int maxn = 5005; 21 struct node { 22 int l, r; 23 int len;//本区间有多长被矩形覆盖 24 int block; 25 bool cl, cr; 26 int cover; 27 }tree[maxn*8]; 28 struct cline { 29 int x, y1, y2; 30 bool left; 31 bool operator <(cline a) { 32 if (x == a.x) 33 return left > a.left; 34 return x < a.x; 35 } 36 }line[maxn*2]; 37 int n,col[maxn*2]; 38 39 void build(int rt, int l, int r) { 40 tree[rt].l = l, tree[rt].r = r; 41 tree[rt].cover = 0,tree[rt].len=0; 42 tree[rt].block = 0, tree[rt].cl = tree[rt].cr = false; 43 if (l == r)return; 44 build(r(rt), mid(l, r) + 1, r); 45 build(l(rt), l, mid(l, r)); 46 } 47 48 void insert(int rt, int l, int r) { 49 if (tree[rt].l == l && tree[rt].r == r) { 50 tree[rt].cover++; 51 tree[rt].len = col[r + 1] - col[l]; 52 tree[rt].block = 1, tree[rt].cl = tree[rt].cr = true; 53 return; 54 } 55 int midn = mid(tree[rt].l, tree[rt].r); 56 if (l > midn) 57 insert(r(rt), l, r); 58 else if (r <= midn) 59 insert(l(rt), l, r); 60 else { 61 insert(l(rt), l, midn); 62 insert(r(rt), midn + 1, r); 63 } 64 if (tree[rt].cover == 0) 65 { 66 tree[rt].len = tree[l(rt)].len + tree[r(rt)].len; 67 tree[rt].cl = tree[l(rt)].cl; 68 tree[rt].cr = tree[r(rt)].cr; 69 tree[rt].block = tree[l(rt)].block + tree[r(rt)].block - tree[l(rt)].cr*tree[r(rt)].cl; 70 } 71 return; 72 } 73 74 void Delete(int rt, int l, int r) { 75 if (tree[rt].l == l && tree[rt].r == r) { 76 tree[rt].cover--; 77 if (tree[rt].cover == 0) { 78 if (l == r) 79 { 80 tree[rt].len = 0; 81 tree[rt].block = 0; 82 tree[rt].cl = tree[rt].cr = false; 83 } 84 else 85 { 86 tree[rt].len = tree[l(rt)].len + tree[r(rt)].len; 87 tree[rt].cl = tree[l(rt)].cl; 88 tree[rt].cr = tree[r(rt)].cr; 89 tree[rt].block = tree[l(rt)].block + tree[r(rt)].block - tree[l(rt)].cr*tree[r(rt)].cl; 90 } 91 } 92 return; 93 } 94 int midn = mid(tree[rt].l, tree[rt].r); 95 if (l > midn) 96 Delete(r(rt), l, r); 97 else if (r <= midn) 98 Delete(l(rt), l, r); 99 else 100 Delete(l(rt), l, midn), Delete(r(rt), midn + 1, r); 101 if (tree[rt].cover == 0) 102 { 103 tree[rt].len = tree[l(rt)].len + tree[r(rt)].len; 104 tree[rt].cl = tree[l(rt)].cl; 105 tree[rt].cr = tree[r(rt)].cr; 106 tree[rt].block = tree[l(rt)].block + tree[r(rt)].block - tree[l(rt)].cr*tree[r(rt)].cl; 107 } 108 return; 109 } 110 111 int binary_find(int s, int e, int val) { 112 while (s < e) { 113 int x = s + (e - s) / 2; 114 if (col[x] >= val)e = x; 115 else s = x + 1; 116 } 117 return s; 118 } 119 120 void init() { 121 scanf("%d", &n); 122 int yc, lc; 123 yc=lc = 0; 124 for (int i = 1; i <= n; i++) { 125 int x1, x2, y1, y2; 126 scanf("%d%d%d%d", &x1, &y1, &x2, &y2); 127 col[++yc] = y1, col[++yc] = y2;; 128 line[++lc].x = x1, line[lc].y1 = y1, line[lc].y2 = y2, line[lc].left = true; 129 line[++lc].x = x2, line[lc].y1 = y1, line[lc].y2 = y2, line[lc].left = false; 130 } 131 sort(col + 1, col + yc+1); 132 yc = unique(col + 1, col + yc + 1) - col - 1;//横线yc条 133 sort(line + 1, line + 1 + lc); 134 build(1, 1, yc - 1); 135 int prec = 0,nowc,ans=0; 136 for (int i = 1; i <= lc; i++) { 137 int l = binary_find(1, yc+1, line[i].y1); 138 int r = binary_find(1, yc+1, line[i].y2)-1; 139 if (line[i].left)insert(1, l, r); 140 else Delete(1, l, r); 141 nowc=tree[1].len; 142 ans += abs(nowc - prec); 143 if(i!=lc) 144 ans+=(tree[1].block * 2 * (line[i + 1].x - line[i].x)); 145 prec = nowc; 146 } 147 printf("%d\n", ans); 148 } 149 150 int main() 151 { 152 init(); 153 return 0; 154 }
我居然以为这道题就是atlantis……
所以我是根据面积并的代码改的周长并……所以体积庞大并且乱七八糟
很迷的是，一开始我提交OpenJudge结果WA了……试着提交POJ却是AC……过了好久我实在改不出啥，几乎没动程序又交了一遍……AC了……
这道题真是折煞我了|||不仅看错题还经历了这种灵异现象
思路：
和atlantis一脉相承
不同在于结构中要加几个元素（为了算横线的长）
还有就是周长的算法变为了：竖线长度和每次需要加上的值是现在 tree[1].len 的值减去上次扫描时同名元素的值（也就是新出现的裸露的线段的值）