HDU 1542 传送门：Problem - 1542
题目大意就是几个矩形相交，然后求出总的覆盖面积，覆盖多次的按一次算。典型的面积并问题。
大体思路：
记录所给的点，排序，去掉重复的点，对点编号，可以理解成存放点的数组的下表即为编号。
对Y轴建立线段树，利用线段树计算面积。
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 #include <cstdlib>
6
7 using namespace std;
8
9 double x[2200],y[2200];
10
11 double st[2200][8000];
12
13 struct N1
14 {
15 double x1,x2,y1,y2;
16 }a[1100];
17
18 int del(double *t,int n)// 去掉重复点的函数
19 {
20 double s[2200];
21 int top = 0,i;
22 for(i = 0;i < n; i++)
23 {
24 if(top == 0)
25 {
26 s[top++] = t[i];
27 }
28 else if(t[i] != s[top-1])
29 {
30 s[top++] = t[i];
31 }
32 }
33 for(i = 0;i < top; i++)
34 {
35 t[i] = s[i];
36 }
37 return top;
38 }
39
40 int check_site(double *t,int s,int e,double x)//二分查找当前点的对应编号
41 {
42 if(s == e)
43 return s;
44 int m = (s+e)/2;
45 if(x == t[m])
46 return m;
47 if(x < t[m])
48 return check_site(t,s,m-1,x);
49 else return check_site(t,m+1,e,x);
50 }
51
52 double change(double *st,int l,int r,int node,int ml,int mr,int site)//对线段树更新
53 {
54 if(ml == l && mr == r && ml+1 == mr)
55 {
56 double area = (y[site+1] - y[site]) * (x[r] - x[l]);
57
58 if(area > st[node])
59 {
60 double temp = area - st[node];
61 st[node] = area;
62 return (temp);
63 }
64 else return 0;
65 }
66
67 int m = (l+r)/2;
68
69 if(mr <= m)
70 {
71 double area = change(st,l,m,node+node,ml,mr,site);
72 st[node] += area;
73 return area;
74 }
75 if(m <= ml)
76 {
77 double area = change(st,m,r,node+node+1,ml,mr,site);
78 st[node] += area;
79 return area;
80 }
81
82 double area = change(st,l,m,node+node,ml,m,site)+change(st,m,r,node+node+1,m,mr,site);
83 st[node] += area;
84 return area;
85 }
86
87 int main()
88 {
89 int n,icase = 1;
90 int i,j;
91 double area;
92
93 while(cin>>n && n)
94 {
95 for(i = 0,j = 0; i < n; i++)
96 {
97 cin>>a[i].x1>>a[i].y1>>a[i].x2>>a[i].y2;
98 x[j] = a[i].x1;
99 y[j] = a[i].y1;
100 j++;
101 x[j] = a[i].x2;
102 y[j] = a[i].y2;
103 j++;
104 }
105
106 sort(x,x+j);
107 int sx = del(x,j);
108 sort(y,y+j);
109 int sy = del(y,j);//排序去掉重复点
110
111 for(i = 0;i <= sy; i++)//初始化线段树
112 {
113 memset(st[i],0,sizeof(st[i]));
114 }
115
116 for(area = 0,i = 0;i < n; i++)
117 {
118 int site_x1 = check_site(x,0,sx-1,a[i].x1);//查找对应编号
119 int site_y1 = check_site(y,0,sy-1,a[i].y1);
120 int site_x2 = check_site(x,0,sx-1,a[i].x2);
121 int site_y2 = check_site(y,0,sy-1,a[i].y2);
122
123 int temp;
124 if(site_x1 > site_x2)
125 {
126 temp = site_x1;
127 site_x1 = site_x2;
128 site_x2 = temp;
129 }
130
131 if(site_y1 > site_y2)
132 {
133 temp = site_y1;
134 site_y1 = site_y2;
135 site_y2 = temp;
136 }
137
138 for(j = site_y1;j < site_y2; j++)//每一棵线段树的第一个节点即为该段区域内的面积
139 {
140 area += change(st[j],0,sx-1,1,site_x1,site_x2,j);//area记录总面积
141 }
142 }
143
144 printf("Test case #%d\n",icase++);
145 printf("Total explored area: %.2lf\n\n",area);
146 }
147 return 0;
148 }
换成这种矩阵模拟的办法也能过 时间和内存都有所减少 瞬间感觉自己丝毫没有窥探到线段树的精髓……sad
矩阵模拟的AC_code:
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 #include <cstdlib>
6
7 using namespace std;
8
9 double x[2200],y[2200];
10
11 double st[2200][8000];
12
13 struct N1
14 {
15 double x1,x2,y1,y2;
16 }a[1100];
17
18 int del(double *t,int n)// 去掉重复点的函数
19 {
20 double s[2200];
21 int top = 0,i;
22 for(i = 0;i < n; i++)
23 {
24 if(top == 0)
25 {
26 s[top++] = t[i];
27 }
28 else if(t[i] != s[top-1])
29 {
30 s[top++] = t[i];
31 }
32 }
33 for(i = 0;i < top; i++)
34 {
35 t[i] = s[i];
36 }
37 return top;
38 }
39
40 int check_site(double *t,int s,int e,double x)//二分查找当前点的对应编号
41 {
42 if(s == e)
43 return s;
44 int m = (s+e)/2;
45 if(x == t[m])
46 return m;
47 if(x < t[m])
48 return check_site(t,s,m-1,x);
49 else return check_site(t,m+1,e,x);
50 }
51
52 /*double change(double *st,int l,int r,int node,int ml,int mr,int site)//对线段树更新
53 {
54 if(ml == l && mr == r && ml+1 == mr)
55 {
56 double area = (y[site+1] - y[site]) * (x[r] - x[l]);
57
58 if(area > st[node])
59 {
60 double temp = area - st[node];
61 st[node] = area;
62 return (temp);
63 }
64 else return 0;
65 }
66
67 int m = (l+r)/2;
68
69 if(mr <= m)
70 {
71 double area = change(st,l,m,node+node,ml,mr,site);
72 st[node] += area;
73 return area;
74 }
75 if(m <= ml)
76 {
77 double area = change(st,m,r,node+node+1,ml,mr,site);
78 st[node] += area;
79 return area;
80 }
81
82 double area = change(st,l,m,node+node,ml,m,site)+change(st,m,r,node+node+1,m,mr,site);
83 st[node] += area;
84 return area;
85 }*/
86
87 double change(int x1,int x2,int y1,int y2)
88 {
89 int i,j;
90 double area = 0;
91 for(i = y1;i < y2; i++)
92 {
93 for(j = x1;j < x2; j++)
94 {
95 if(st[i][j] == 0)
96 {
97 st[i][j] = (x[j+1]-x[j]) * (y[i+1]-y[i]);
98 area += st[i][j];
99 }
100 }
101 }
102 return area;
103 }
104
105 int main()
106 {
107 int n,icase = 1;
108 int i,j;
109 double area;
110
111 while(cin>>n && n)
112 {
113 for(i = 0,j = 0; i < n; i++)
114 {
115 cin>>a[i].x1>>a[i].y1>>a[i].x2>>a[i].y2;
116 x[j] = a[i].x1;
117 y[j] = a[i].y1;
118 j++;
119 x[j] = a[i].x2;
120 y[j] = a[i].y2;
121 j++;
122 }
123
124 sort(x,x+j);
125 int sx = del(x,j);
126 sort(y,y+j);
127 int sy = del(y,j);//排序去掉重复点
128
129 for(i = 0;i <= sy; i++)//初始化线段树
130 {
131 memset(st[i],0,sizeof(st[i]));
132 }
133
134 for(area = 0,i = 0;i < n; i++)
135 {
136 int site_x1 = check_site(x,0,sx-1,a[i].x1);//查找对应编号
137 int site_y1 = check_site(y,0,sy-1,a[i].y1);
138 int site_x2 = check_site(x,0,sx-1,a[i].x2);
139 int site_y2 = check_site(y,0,sy-1,a[i].y2);
140
141 int temp;
142 if(site_x1 > site_x2)
143 {
144 temp = site_x1;
145 site_x1 = site_x2;
146 site_x2 = temp;
147 }
148
149 if(site_y1 > site_y2)
150 {
151 temp = site_y1;
152 site_y1 = site_y2;
153 site_y2 = temp;
154 }
155
156 area += change(site_x1,site_x2,site_y1,site_y2);
157
158 /*for(j = site_y1;j < site_y2; j++)//每一棵线段树的第一个节点即为该段区域内的面积
159 {
160 area += change(st[j],0,sx-1,1,site_x1,site_x2,j);//area记录总面积
161 }*/
162 }
163
164 printf("Test case #%d\n",icase++);
165 printf("Total explored area: %.2lf\n\n",area);
166 }
167 return 0;
168 }
