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.
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.
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
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 ",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 }