Hdu 1255
覆盖的面积
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3545 Accepted Submission(s): 1739Problem Description给定平面上若干矩形,求出被这些矩形覆盖过至少两次的区域的面积.
Input输入数据的第一行是一个正整数T(1<=T<=100),代表测试数据的数量.每个测试数据的第一行是一个正整数N(1<=N<=1000),代表矩形的数量,然后是N行数据,每一行包含四个浮点数,代表平面上的一个矩形的左上角坐标和右下角坐标,矩形的上下边和X轴平行,左右边和Y轴平行.坐标的范围从0到100000.
注意:本题的输入数据较多,推荐使用scanf读入数据.
Output对于每组测试数据,请计算出被这些矩形覆盖过至少两次的区域的面积.结果保留两位小数.
Sample Input2 5 1 1 4 2 1 3 3 7 2 1.5 5 4.5 3.5 1.25 7.5 4 6 3 10 7 3 0 0 1 1 1 0 2 1 2 0 3 1
Sample Output7.63 0.00
计算被两个及以上矩形覆盖的面积和。
此种类型题目有两种做法。
第一种效率较低：
Accepted Code:
1 /************************************************************************* 2 > File Name: G.cpp 3 > Author: Stomach_ache 4 > Mail: [email protected] 5 > Created Time: 2014年07月26日 星期六 19时05分05秒 6 > Propose: hdu 1542 7 ************************************************************************/ 8 #include <cmath> 9 #include <string> 10 #include <cstdio> 11 #include <fstream> 12 #include <cstring> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 17 const int maxn = 1100; 18 struct Line { 19 double x, y1, y2; 20 int cover; 21 friend bool operator < (Line a, Line b) { 22 return a.x < b.x; 23 } 24 }line[maxn*2 + 5]; 25 26 struct node { 27 double x, y1, y2; 28 int cover, flag; 29 }Tree[maxn*1000]; 30 int n; 31 double y[maxn*2]; 32 33 void build(int root, int L, int R) { 34 Tree[root].x = -1; 35 Tree[root].cover = 0; 36 Tree[root].y1 = y[L]; 37 Tree[root].y2 = y[R]; 38 Tree[root].flag = 0; 39 if (R - L == 1) { 40 Tree[root].flag = 1; 41 return ; 42 } 43 int M = (L + R) / 2; 44 build(root * 2, L, M); 45 build(root * 2 + 1, M, R); 46 } 47 48 double insert(int root, Line &ll) { 49 if (Tree[root].y1 >= ll.y2 || Tree[root].y2 <= ll.y1) { 50 return 0.0; 51 } 52 if (Tree[root].flag) { 53 if (Tree[root].cover > 1) { 54 double ans = (ll.x - Tree[root].x) * (Tree[root].y2 - Tree[root].y1); 55 Tree[root].x = ll.x; 56 Tree[root].cover += ll.cover; 57 return ans; 58 } else { 59 Tree[root].cover += ll.cover; 60 Tree[root].x = ll.x; 61 return 0.0; 62 } 63 } 64 double ans1 = insert(root * 2, ll); 65 double ans2 = insert(root * 2 + 1, ll); 66 return ans1 + ans2; 67 } 68 69 int main(void) { 70 #ifndef ONLINE_JUDGE 71 freopen("in.txt", "r", stdin); 72 #endif 73 int cas; 74 scanf("%d", &cas); 75 while (cas--) { 76 scanf("%d", &n); 77 int cnt = 0; 78 for (int i = 0; i < n; i++) { 79 double xx, yy, xxx, yyy; 80 scanf("%lf %lf %lf %lf", &xx, &yy, &xxx, &yyy); 81 y[++cnt] = yy; 82 line[cnt].x = xx; line[cnt].y1 = yy; line[cnt].y2 = yyy; 83 line[cnt].cover = 1; 84 y[++cnt] = yyy; 85 line[cnt].x = xxx; line[cnt].y1 = yy; line[cnt].y2 = yyy; 86 line[cnt].cover = -1; 87 } 88 sort(y + 1, y + cnt + 1); 89 int len = 1; 90 for (int i = 2; i <= cnt; i++) { 91 if (y[i] != y[len]) { 92 y[++len] = y[i]; 93 } 94 } 95 sort(line + 1, line + cnt + 1); 96 build(1, 1, len); 97 double ans = 0.0; 98 for (int i = 1; i <= cnt; i++) { 99 ans += insert(1, line[i]); 100 } 101 printf("%.2f\n", ans); 102 } 103 104 return 0; 105 }
第二种做法效率较高。
Accepted Code:
1 /************************************************************************* 2 > File Name: G.cpp 3 > Author: Stomach_ache 4 > Mail: [email protected] 5 > Created Time: 2014年07月26日 星期六 19时05分05秒 6 > Propose: hdu 7 ************************************************************************/
8 #include <map>
9 #include <cmath>
10 #include <string>
11 #include <cstdio>
12 #include <vector>
13 #include <fstream>
14 #include <cstring>
15 #include <iostream>
16 #include <algorithm>
17 using namespace std; 18
19 const int maxn = 1100; 20 //保存每条线段的信息
21 struct Line { 22 double x, y1, y2; 23 //矩形的左边或者右边
24 int flag; 25 friend bool operator < (Line a, Line b) { 26 return a.x < b.x; 27 } 28 }line[maxn*2 + 5]; 29
30 //线段树结点信息
31 struct node { 32 int l, r; 33 //线段被覆盖的次数
34 int cover; 35 //len为被覆盖一次的长度，len2为被覆盖两次的长度
36 double len, len2; 37 }Tree[maxn*40]; 38 int n; 39 //保存所有y轴坐标
40 double y[maxn*2]; 41 //离散化使用
42 vector<double> xs; 43
44 void build(int root, int L, int R) { 45 Tree[root].l = L; 46 Tree[root].r = R; 47 Tree[root].len = 0.0; 48 Tree[root].len2 = 0.0; 49 Tree[root].cover = 0; 50 if (R - L == 1) { 51 return ; 52 } 53 int M = (L + R) / 2; 54 build(root * 2, L, M); 55 build(root * 2 + 1, M, R); 56 } 57
58 void pushup(int root) { 59 //被覆盖两次以上
60 if (Tree[root].cover > 1) { 61 Tree[root].len = 0; 62 Tree[root].len2 = xs[Tree[root].r-1] - xs[Tree[root].l-1]; 63 } else if (Tree[root].cover == 1) { //被覆盖一次
64 Tree[root].len2 = Tree[root*2].len + Tree[root*2+1].len 65 +Tree[root*2].len2 + Tree[root*2+1].len2; 66 Tree[root].len = xs[Tree[root].r-1] - xs[Tree[root].l-1] 67 - Tree[root].len2; 68 } else { //没有被覆盖
69 if (Tree[root].l + 1 == Tree[root].r) 70 Tree[root].len = Tree[root].len2 = 0.0; 71 else { 72 Tree[root].len = Tree[root*2].len + Tree[root*2+1].len; 73 Tree[root].len2 = Tree[root*2].len2 + Tree[root*2+1].len2; 74 } 75 } 76 } 77
78 void update(int root, int L, int R, int flag) { 79 //不在当前区间内
80 if (Tree[root].l >= R || Tree[root].r <= L) 81 return ; 82 //包含在当前区间内
83 if (Tree[root].l >= L && Tree[root].r <= R) { 84 Tree[root].cover += flag; 85 pushup(root); 86 return ; 87 } 88 int M = (Tree[root].l + Tree[root].r) / 2; 89 update(root*2, L, R, flag); 90 update(root*2+1, L, R, flag); 91 pushup(root); 92 } 93
94 //离散化
95 int compress(int m) { 96 xs.clear(); 97 for (int i = 1; i <= m; i++) xs.push_back(y[i]); 98 sort(xs.begin(), xs.end()); 99 xs.erase(unique(xs.begin(), xs.end()), xs.end()); 100 return xs.size(); 101 } 102
103
104 int main(void) { 105 #ifndef ONLINE_JUDGE 106 freopen("in.txt", "r", stdin); 107 #endif
108 int cas; 109 scanf("%d", &cas); 110 while (cas--) { 111 scanf("%d", &n); 112 int cnt = 0; 113 for (int i = 0; i < n; i++) { 114 double xx, yy, xxx, yyy; 115 scanf("%lf %lf %lf %lf", &xx, &yy, &xxx, &yyy); 116 y[++cnt] = yy; 117 line[cnt].x = xx; line[cnt].y1 = yy; line[cnt].y2 = yyy; 118 line[cnt].flag = 1; 119 y[++cnt] = yyy; 120 line[cnt].x = xxx; line[cnt].y1 = yy; line[cnt].y2 = yyy; 121 line[cnt].flag = -1; 122 } 123 int w = compress(cnt); 124 sort(line + 1, line + cnt + 1); 125 build(1, 1, cnt); 126 double ans = 0.0; 127 int l = find(xs.begin(), xs.end(), line[1].y1) - xs.begin() + 1; 128 int r = find(xs.begin(), xs.end(), line[1].y2) - xs.begin() + 1; 129 update(1, l, r, line[1].flag); 130 for (int i = 2; i <= cnt; i++) { 131 int l = find(xs.begin(), xs.end(), line[i].y1) - xs.begin() + 1; 132 int r = find(xs.begin(), xs.end(), line[i].y2) - xs.begin() + 1; 133 ans += (line[i].x - line[i-1].x) * Tree[1].len2; 134 update(1, l, r, line[i].flag); 135 } 136 printf("%.2f\n", ans); 137 } 138
139 return 0; 140 }