标签：des style blog http java color 使用 os
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
1 /************************************************************************* 2 > File Name: G.cpp 3 > Author: Stomach_ache 4 > Mail: sudaweitong@gmail.com 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。
1 /*************************************************************************
2 > File Name: G.cpp
3 > Author: Stomach_ache
4 > Mail: sudaweitong@gmail.com
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 }
标签：des style blog http java color 使用 os
原文地址：http://www.cnblogs.com/Stomach-ache/p/3871741.html
