http://acm.hdu.edu.cn/showproblem.php?pid=1828
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Problem 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.
Please process to the end of file.
0 <= number of rectangles < 5000
All coordinates are in the range [-10000,10000] and any existing rectangle has a positive area.
Please process to the end of file.
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
Meaning of the questions:
A plurality of rectangular, parallel to the two sides of the rectangular coordinate axes, between the rectangles may cover each other, find the circumference.
Ideas:
Each record two vertical side of the rectangle (x1, y1, y2) and (x2, y1, y2), all in accordance with the vertical side x from small to large, then a start to a vertical edge perimeter, so to vertical vertical side where the x-axis, i.e., a straight line is a scanning line.
Each move to a new vertical side, we need to calculate the scanning line is useful where the vertical side edge length (i.e., the vertical side of the useful part of the current, the current may be covered with the vertical side portion), and added to the current scan line and before the scan line length * number of lateral transverse edge strip,
It has been calculated to the last vertical edge, that is, the complete perimeter.
code show as below:
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <string> 5 #include <math.h> 6 #include <algorithm> 7 #include <vector> 8 #include <queue> 9 #include <set> 10 #include <stack> 11 #include <map> 12 #include <math.h> 13 const int INF=0x3f3f3f3f; 14 typedef long long LL; 15 const int mod=1e9+7; 16 //const double PI=acos(-1); 17 const int maxn=1e5+10; 18 using namespace std; 19 //ios::sync_with_stdio(false); 20 // cin.tie(NULL); 21 22 const int N=5005; 23 24 struct Line_node 25 { 26 int x; 27 int y1,y2; 28 int flag; 29 BOOL operator <( const Line_node & S) 30 { 31 is IF (X == SX) 32 return In Flag> S.FLAG; 33 is the else 34 is return X < SX; 35 } 36 } Line [N * 2 ]; 37 [ 38 is struct SegTree_node 39 { 40 int L; 41 is int R & lt; 42 is BOOL LF, RF; // about whether the boundary point is covered; 43 is int cover_len; 44 int cover_num; 45 int num;//矩形数目 46 }SegTree[maxn<<2]; 47 48 vector<int> vt; 49 50 void Build(int l,int r,int rt) 51 { 52 SegTree[rt].l=l; 53 SegTree[rt].r=r; 54 SegTree[rt].cover_len=0; 55 SegTree[rt].cover_num=0; 56 SegTree[rt].num=0; 57 SegTree[rt].lf=SegTree[rt].rf=false; 58 if(l+1==r) 59 return ; 60 int mid=(l+r)>>1; 61 Build(l,mid,rt<<1); 62 Build(mid,r,rt<<1|1); 63 } 64 65 void PushUp(int rt) 66 { 67 int l=SegTree[rt].l; 68 intr = SegTree [rt] .r; 69 if (SegTree [rt] .cover_num> 0 ) 70 { 71 SegTree [rt] .cover_len = vt [r] - vt [l]; 72 SegTree [rt] = .lf SegTree [rt] .rf = true ; 73 SegTree [rt] .num = 1 ; 74 return ; 75 } 76 // if (l + 1 == r) 77 // { 78 // SegTree [rt] .cover_len = 0; 79 // SegTree [rt] = .lf SegTree [rt] .rf = false; 80 // SegTree [rt] .num = 0; 81 // return; 82 // } 83 SegTree [rt] .cover_len SegTree = [rt << 1 ] + .cover_len SegTree [rt << 1 | 1 ] .cover_len; 84 SegTree [RT] = .num SegTree [rt << 1 ] + .num SegTree [rt << 1 | 1 ] .num- (SegTree [rt << 1 ] & .rf SegTree [rt << 1 | 1 ] .lf) 85 SegTree [rt] .lf SegTree = [rt << 1 ] .lf; 86 SegTree [RT] = .rf SegTree [rt << 1 | 1 ] .rf; } 88 89 void Update(Line_node t,int rt) 90 { 91 int l=SegTree[rt].l; 92 int r=SegTree[rt].r; 93 if(t.y1<=vt[l]&&t.y2>=vt[r]) 94 { 95 SegTree[rt].cover_num+=t.flag; 96 PushUp(rt); 97 return ; 98 } 99 int mid=(l+r)>>1; 100 if(t.y1<vt[mid]) 101 Update(t,rt<<1); 102 if(t.y2>vt[mid]) 103 Update(t,rt<<1|1); 104 PushUp(rt); 105 } 106 107 int main() 108 { 109 int n; 110 while (~scanf("%d",&n)) 111 { 112 vt.clear(); 113 for(int i=0;i<n;i++) 114 { 115 int x1,x2,y1,y2; 116 scanf("%d %d %d %d",&x1,&y1,&x2,&y2); 117 Line[i*2].x=x1; 118 Line[i*2].y1=y1; 119 Line[i*2].y2=y2; 120 Line[i*2].flag=1; 121 122 Line[i*2+1].x=x2; 123 Line[i*2+1].y1=y1; 124 Line [I * 2 + . 1 ] .Y2 = Y2; 125 Line [I * 2 + . 1 ] .flag = - . 1 ; 126 vt.push_back (Y1); 127 vt.push_back (Y2); 128 } 129 Sort (Line , Line + 2 * n-); 130. // Y coordinate discretized 131 is Sort (vt.begin (), vt.end ()); 132 int NUM = UNIQUE (vt.begin (), vt.end ()) - VT .begin (); // to obtain discrete finished weight and number 133 the Build ( 0 , num- . 1 ,1); 134 int ans=0; 135 int prelen=0; 136 for(int i=0;i<n*2;i++) 137 { 138 if(i>0) 139 { 140 ans+=SegTree[1].num*2*(Line[i].x-Line[i-1].x); 141 } 142 Update(Line[i],1); 143 ans + = abs (SegTree [ 1 ] .cover_len- prelen); 144 prelen SegTree = [ 1 ] .cover_len; 145 } 146 printf ( " % d \ n " , ans); 147 } 148 return 0 ; 149 }