面积并问题 HDU 1542 Atlantis

HDU 1542 传送门：http://acm.hdu.edu.cn/showproblem.php?pid=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。

　　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 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 }

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　　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  }

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