POJ1177 Picture

Posted 2020-08-19

Description

A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others. The length of the boundary of the union of all rectangles is called the perimeter. 

Write a program to calculate the perimeter. An example with 7 rectangles is shown in Figure 1. 

The corresponding boundary is the whole set of line segments drawn in Figure 2. 

The vertices of all rectangles have integer coordinates. 

Input

Your program is to read from standard input. The first line contains the number of rectangles pasted on the wall. In each of the subsequent lines, one can find the integer coordinates of the lower left vertex and the upper right vertex of each rectangle. The values of those coordinates are given as ordered pairs consisting of an x-coordinate followed by a y-coordinate. 

0 <= number of rectangles < 5000 
All coordinates are in the range [-10000,10000] and any existing rectangle has a positive area.

Output

Your program is to write to standard output. The output must contain a single line with a non-negative integer which corresponds to the perimeter for the input rectangles.

Sample Input

7
-15 0 5 10
-5 8 20 25
15 -4 24 14
0 -6 16 4
2 15 10 22
30 10 36 20
34 0 40 16

Sample Output

228

Source

IOI 1998

大概就是线段树+离散化搞搞就好了，之前一直不会线段树的离散化，大概就是边界总是要弄错，然后做了这道题，大概就知道了，可以多维护两个量表示左右是否覆盖的情况。

  1 #include <algorithm>
  2 #include <iostream>
  3 #include <cstdlib>
  4 #include <cstdio>
  5 const int N = 10000 + 11 ;
  6 using namespace std;
  7 int n ,li[N] , tot;
  8 struct id
  9 {
 10     int x,y1,x2,y2,flag;
 11 } line[N];
 12 struct a_seg_tree
 13 {
 14     struct a_tree
 15     {
 16         int l,r,cnt,qucnt,len,lenall,seg,mid;
 17         bool rbd,lbd ;
 18     } tree[N<<2];
 19     void build(int l,int r,int num) 
 20     {
 21         tree[num].l = l , tree[num].r = r;
 22         tree[num].lenall = tree[num].cnt = tree[num].seg = 0;
 23         int mid = l + ( (r - l )>>1),ii = num << 1 ;
 24         tree[num].len = li[r] - li[l];
 25         tree[num].mid = mid;
 26         if(r - l > 1)
 27         {
 28             build(l,mid,ii);
 29             build(mid,r,ii+1);
 30         }
 31     }
 32     void Update(int num)
 33     {
 34         if(tree[num].cnt > 0)
 35         {
 36             tree[num].lenall = tree[num].len;
 37             tree[num].seg = 1; 
 38             tree[num].lbd = tree[num].rbd = true ;
 39         }
 40         else if(tree[num].r-tree[num].l>1)
 41         {
 42             int ii = num << 1;
 43             tree[num].lenall = tree[ii].lenall + tree[ii+1].lenall;
 44             tree[num].lbd = tree[ii].lbd;
 45             tree[num].rbd = tree[ii+1].rbd;
 46             tree[num].seg = tree[ii].seg + tree[ii+1].seg - (tree[ii].rbd & tree[ii+1].lbd);
 47         }
 48         else
 49         {
 50             tree[num].lenall = tree[num].seg = 0;
 51             tree[num].lbd = tree[num].rbd = false;
 52         }
 53         
 54     }
 55     void Insert(int l,int r,int num)
 56     {
 57         if(l == tree[num].l && r == tree[num].r) ++tree[num].cnt;
 58         else
 59         {
 60             int mid = tree[num].mid,ii = num << 1;
 61             if(r <= mid) Insert(l,r,ii);
 62             else if(l >= mid) Insert(l,r,ii+1);
 63             else Insert(l,mid,ii),Insert(mid,r,ii+1);
 64         }
 65         Update(num);
 66     }
 67     void Delete(int l,int r,int num)
 68     {
 69         if(l == tree[num].l && r == tree[num].r)--tree[num].cnt;
 70         else
 71         {
 72             int mid = tree[num].mid,ii = num << 1;
 73             if(r <= mid)Delete(l,r,ii);
 74             else if(l>=mid)Delete(l,r,ii+1);
 75             else Delete(l,mid,ii),Delete(mid,r,ii+1);
 76         }
 77         Update(num);
 78     }
 79 }seg_tree;
 80 
 81 int cmp(id a,id b)
 82 {
 83     if(a.x == b.x)return a.flag > b.flag;
 84     return a.x < b.x;
 85 }
 86 
 87 
 88 void Init()
 89 {
 90     scanf("%d",&n);
 91     int x1,y1,x2,y2,i= 0;
 92     while(n--)
 93     {
 94         scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
 95         line[++i].x=x1,line[i].y1=y1,line[i].y2=y2,line[i].flag=1;li[i]=y1;
 96         line[++i].x=x2,line[i].y1=y1,line[i].y2=y2,line[i].flag=-1;li[i]=y2;
 97     }
 98     sort(li+1,li+1+i); n = i ;
 99     sort(line+1,line+1+i,cmp);
100     tot = 1 ;
101     for(int j = 2; j <= i ; ++j )
102         if(li[j] != li[tot])li[++tot] = li[j];
103 }
104 
105 int binary_search( int sum )
106 {
107     int l = 1 , r = tot;
108     while(l <= r)
109     {
110         int mid = l + ((r-l)>>1);
111         if(li[mid] == sum)return mid;
112         if(sum<li[mid])r = mid - 1;
113         else l = mid + 1;
114     }
115     return -1;
116 }
117 
118 void Solve()
119 {
120     seg_tree.build(1,tot,1);
121     int ans = 0 , pre = 0 ; 
122     for(int i = 1; i < n;++i)
123     {
124         int y1 = binary_search(line[i].y1) , y2 = binary_search(line[i].y2) ;    
125         if(line[i].flag == 1 )
126             seg_tree.Insert(y1,y2,1) ; 
127         else
128             seg_tree.Delete(y1,y2,1);
129         ans += seg_tree.tree[1].seg * (line[i+1].x-line[i].x) << 1 ;
130         ans += abs(seg_tree.tree[1].lenall-pre);
131         
132         pre = seg_tree.tree[1].lenall;
133     }
134     seg_tree.Delete(binary_search(line[n].y1),binary_search(line[n].y2),1);
135     ans += abs(seg_tree.tree[1].lenall - pre) ;
136     printf("%d\n",ans);
137 }
138 
139 int main()
140 {
141 //    freopen("picture.in","r",stdin);
142 //    freopen("picture.out","w",stdout);
143     Init();
144     Solve();
145     fclose(stdin);
146     fclose(stdout);
147     return 0 ;    
148 }