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 Input
2
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 1Sample Output
7.63
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 }
时间： 2024-10-13 14:54:37